Page 1
1 1
SIMPLE AND COMPOUND INTEREST
Unit Structure
1.0 Objective
1.1 Introduction
1.2 Simple interest
1.3 Compound interest
1.4 Effective rates of interest
1.5 Let us sum up
1.6 Unit end exercise
1.7 List of References
1.0 OBJECTIVES
After going through t his chapter you will able to know:
Calculation of simple interest.
Calculation of compound interest.
Difference between nominal and effective rate of interest.
1.1 INTRODUCTION
We have to work with money everyday life. When a borrower borrows
money from a lender or any financial institution or bank, there is some
extra amount that is charged on the total amount that is borrowed. This
extra amount is termed as an interest rate. Interest charged can be of two
type’s simple interest and compound interest.
To summarize, the concept of simple interest is the amount paid for the
money borrowed for a fixed period of time. While in the case of
compound interest, whenever the interest isup for payment, it is added
back to the principal amount. In this chapter we wil l highlight the
differences between simple interest and compound interest.
1.2 SIMPLE INTEREST
Principal : The sum borrowed by a person is called its principal . It is
denoted by P.
Period : The time span for which money is lent is called period . It is
denot ed by n.
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Business
Mathematics
2 Interest : The amount paid by a borrower to the lender for the use of
money borrowed for a certain period of time is called Interest . It is
denoted by I.
Rate of Interest : This is the interest to be paid on the amount of Rs. 100
per annum (i.e. per year). This is denoted by r.
Total Amount : The sum of the principal and interest is called as the total
amount and is denoted by A. Thus, A = P + I .
i.e. Interest paid I = A – P.
Simple Interest :
The interest which is payable on the principal only is called as simple
interest (S.I.). For example the interest on Rs. 100 at 11% after one year is
Rs.11 and the amount is 100 + 11 = Rs. 111.
It is calculated by the formula: I = 100Pnr
Amount at the end of nth year = A = P + I = P + 100Pnr= P1100nr
Example 1: If Mr. Amol borrows Rs. 1000 for 4 years at 5% rate of
interest, find (i) simple interest and (ii) total amount.
Answer: Given P = Rs. 1000, n = 4 and r = 5
(i) I = 100Pnr= 1000 x 4 x 5100= Rs. 200
(ii) A = P + I = 1000 + 200 = Rs. 1200
Example 2: If Mr. Aasim borrows Rs. 500 for 5 years and pays an interest
of Rs. 175, find rate of interest.
Answer: Given P = 500, n = 5 and I = Rs. 175
Now, I = 100Pnr x 100 x IrPn 175 x 100500 x 5r = 7
Example 3: Find the period for Rs. 2500 to yield Rs. 900 in simple interest
at 12%
Answer: Given P = Rs. 2500, I = 900, r = 12
Now, I = 100Pnr x 100 x InPr 900 x 1002500 x 12n = 3
The period is 3 years.
Example 4: Mr. Akshay lent Rs. 5000 to Mr. Amit and Rs. 4000 to Mr.
Sunil for 5 years and received total simple interest of Rs. 4950. Find (i) the
rate of interest and (ii) simple interest of each.
Answer: Let the rate of interest be r. munotes.in
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Simple and Compound
Interest
3 S.I. for Anil = 5000 x 5 x 100r= 250 r … (1)
and S.I. for Sunil = 4000 x 5 x 100r = 200 r … (2)
from (1) and (2), we have,
total interest from both = 250 r + 200 r
= 450 r
But total interest received be Mr. Akshay = Rs. 4950 450r = 4950 r = 4950450= 11
Example 5: The S.I. on a sum of money is one -fourth the principal. If the
period is same as that of the rate of intere st then find the rate of interest.
Answer: Given I = 4P and n = r
Now, w.k.t. I = 100Pnr 4P= x x 100Prr 1004= r2 r2 = 25 r = 5. The rate of interest = 5%
Example 6: If Rs. 8400 amount to Rs. 11088 in 4 years, what will Rs.
10500 amount to in 5 years at the same rate of interest?
Answer:
(i) Given n = 4, P = Rs. 8400, A = Rs. 11088 I = A – P = 11088 – 8400 = Rs. 2688
Let r be the rate of interest. r = 8%
Now, I = 100Pnr 2688 = 8400 x 4 x 100r
(ii) To find A when n = 5, P = Rs. 10500, r = 8
A = P 1100nr= 10500 x 5 x 81100= 10500 x 140100= 14700 The required amount = Rs. 14,700
the rate of interest = munotes.in
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Business
Mathematics
4 Example 7:Mr. Arvind Kamble earns Rs. 14 00 after 5 years by lending at
a certain rate of interest Rs. 2300 to Mr. Vinod and Rs. 2750 to Mr.
Kuldeep at a rate of interest 1% more than that of Mr. Vinod. Find the
rates of interest.
Answer: Let the rate of interest for Mr. Vinod ber %.
Hence the ra te of interest for Mr. Kuldeep is ( r + 1) %.
S.I. from Mr. Vinod = 2300 x 5 x 100r= 115. r … (1)
S.I. from Mr. Kuldeep = 2750 x 5 x ( 1)100r = 137.5( r + 1) … (2)
The total interest received by Mr. Arvind from both = 115 r + 137.5( r + 1)
= 115 r + 137.5 r + 137.5
= 252.5 r + 137.5 = 1400 … given 252.5 r = 1262.5 The rate of interest for Mr. Vinod = 5% and the rate of interest for Mr.
Kuldeep = 6%
1.3 COMPOUND INTEREST
The interest w hich is calculated on the amount in the previous year is
called compound interest.
For example, the compound interest on Rs. 100 at 8% after one year is Rs.
8 and after two years is 108 + 8% of 108 = Rs. 116.64
If P is the principal, r is the rate of int erest p.a. then the amount at the end
of nth year called as compound amount is given by the formula:
The compound interest is given by the formula:
Note :
1. The interest may be compounded annually (yearly), semi -annually
(half yearly), quarterly or mon thly. Thus, the general formula to calculate
the amount at the end of n years is as follows:
Here p: number of times the interest is compounded in a year.
p = 1 if interest is compounded annually A = 1100nrP
CI = A – P
A = 1 x 100nprPp munotes.in
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Simple and Compound
Interest
5 p = 2 if interest is compounded semi -annually
p = 4 if interest is compounded quarterly
p = 12 if interest is compounded monthly
2. It is easy to calculate amount first and then the compound interest as
compared with finding interest first and then amount in case of simple
interest.
Example 8:Find the compound amount and compound interest of Rs. 800
invested for 10 years at 8% if the interest is compounded annually.
Answer: Given P = 800, r = 8, n = 10.
Since the interest is compounded annually, we have
A = 1100nrP= 800 x 1081100= 800 x 2.1589 ≈ Rs. 1727
Example 9: Find the compound amount and compound interest of Rs.
1000 invested for 5 years at 6% if the interest is compounded ( i) annually,
(ii) semi annually, ( iii) quarterly and ( iv) monthly.
Answer: Given P = Rs. 1000, r = 6, n = 5
(i) If the interest is compounded annually:
A = 1100nrP= 1000 x 561100= 1000 x 1.3382 ≈ Rs. 1338
CI = A – P = 1338 – 1000 = Rs. 338
(ii) If the interest is compounded semi -annually:
A = 212 x 100nrP= 1000 x 1061200= 1000 x 1.3440 = Rs. 1340
CI = A – P = 1340 – 1000 = Rs. 340
(iii) If the interest is compounded quarterly:
A = 414 x 100nrP= 1000 x 2061400= 1000 x 1.347 = Rs. 1347
CI = A – P = 1347 – 1000 = Rs. 347
(iv) If the interest is compounded monthly:
A = 12112 x 100nrP= 1000 x 60611200= 1000 x 1.349 = Rs. 1349
CI = A – P = 1349 – 1000 = Rs. 349
Example 10: Find the principal which will amount to Rs. 11, 236 in 2
years at 6% compound interest compounded annually.
Answer: Given A = Rs. 11236, n = 2, r = 6 and P= ?
Now, A = 1100nrP munotes.in
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Business
Mathematics
6 11236 = P261100= P x 1.1236 P = 112361.1236= 10,000 the required principal is Rs. 10,000.
Example 11:Dr. Ashwinikumar wants to invest some amount for 4 years
in a bank. Bank X offers 8% interest if compounded half yearly while bank
Y offers 6% intere st if compounded monthly. Which bank should Dr.
Ashwinikumar select for better benefits?
Answer: Given n = 4,
Let the principal Mr. Ashwinikumar wants to invest be P = Rs. 100
From Bank X:r = 8 and interest is compounded half -yearly, so p = 2. A = 212 x 100nrP= 100 x 481200= 116.9858 … (1)
From Bank Y: r = 6, p = 12 A = 12112 x 100nrP= 100 x 48611200= 127.0489 … (2)
Comparing (1) and (2) , Dr. Ashwinikumar should invest his amount in
bank Y as it gives more interest at the end of the period.
Example 12: In how many years would Rs. 75,000 amount to Rs.
1,05,794.907 at 7% compound interest compounded semi -annually?
Answer: Given A = Rs. 10 5794.907, P = Rs. 75000, r = 7, p = 2
A = 212 x 100nrP 105794.907 = 75000 x 271200n 105794.90775000= 21.035n 1.410 59876 = (1.035) 2n (1.035) 10 = (1.035) 2n 2n = 10 n = 5
Example 13: A certain principal amounts to Rs. 4410 after 2 years and to
Rs.4630.50 after 3 years at a certain rate of i nterest compounded annually.
Find the principal and the rate of interest.
Answer: Let the principal be P and rate of interest be r.
Now, we know that 1100nrAP
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Simple and Compound
Interest
7 From the given data we have,
4410 = P21100r and 4630.5 = P31100r 4410 = P(1 + 0.01 r) 2 … (1)
4630.5 = P(1 + 0.01 r)3 … (2)
Dividing (2) by (1), we have 4630.54410= 32(1 0.01 )(1 0.01 )PrPr
1.05 = 1 + 0.01 r 0.05 = 0.01 r r = 5
Example 14: Find the rate of interest at which a sum of Rs. 2000 amounts
to Rs. 2690 in 3 years given that the interest is compounded half yearly.
( 61.345 = 1.05)
Answer: Given P = Rs. 2000, A = Rs. 2680, n = 3, p = 2
Now, A = 212 x 100nrP 2690 = 2000 x 61200r 26902000= 61200r 1.345 = 61200r 61.345= 1 + 200r 1.05 = 1 + 200r r = 0.05 x 200 = 10%
Thus, the rate of compo und interest is 10 % .
Example 15: If the interest compounded half yearly on a certain principal
at the end of one year at 8% is Rs. 3264, find the principal.
Answer: Given CI = Rs. 3264, n = 1, p = 2 and r = 8
Now, CI = A – P = 281200P– P
i.e. 3264 = P[ (1.04) 2 – 1] = 0.0816 P P = 32640.0816= 40000
Thus, the principal is Rs. 40,000. munotes.in
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Business
Mathematics
8 1.4 EFFECTIVE RATES OF INTEREST
Let a person invest Rs. 100 at 6% rate of interest p.a. compounded half
yearly. At the end of the year the compound amount is 100(1.03) 2 =
106.09. This means that the annual rate of return is 6.09% which is
actually more than normal (also called nominal) rate of interest 6%. Such
a rate of interest is called as effective rate of intere st. It is denoted by Re.
The formula to compute the effective rate of interest is as follows:
Note :
1. If the period of conversion is annual then the effective rate is same as
the normal rate of interest.
2. Effective rates are used to compare different r ates of interest and
facilitate to find the most beneficial.
Example 16:A bank X offers 8% interest compounded semi -annually
while another bank offers 8.5% interest compounded monthly. Which
bank gives more interest at the end of the year?
Answer: The effe ctive rate for bank X is: 2812 x 100– 1 = (1.04) 2 – 1 = 0.0816 = 8.16% … (1)
While, the effective rate for bank Y is: 128.5112 x 100– 1 = (1.0071) 12 – 1 = 0.0886 = 8.86% … (2)
From (1) and (2), we conclude that the bank Y gives more return at the end
of the year.
Example 17: What is the effective rate equivalent to the normal rate of
10% compounded monthly?
Answer: Given r = 10 and p = 12 Re = 1 x 100pr
p– 1 = 121011200– 1 = 1.1047 – 1 = 0.1047 the effective rate is 10.47%
Re = 1 x 100pr
p– 1
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Simple and Compound
Interest
9 1.5 LET US SUM UP
In this chapter we have learn:
To calculate simple interest.
To calculate compound interest.
To calculate nominal and effective rate of interest.
Differ ence between compound interest.
1.6 UNIT END EXERCISE
1. Find the SI and amount for the following data giving principal, rate of
interest and number of years:
(i) 1800, 6%, 4 years. (ii) 4500, 8%, 5 years
(iii) 7650, 5.5%, 3 years. (iv) 6000, 7.5%, 6 year s
(v) 25000, 8%, 5 years (vi) 20000, 9.5%, 10 years.
2. Find the S.I. and the total amount for a principal of Rs. 6000 for 3
years at 6% rate of interest.
3. Find the S.I. and the total amount for a principal of Rs. 11000 for 5
years at 4% rate of interest.
4. Find the S.I. and the total amount for a principal of Rs. 3300 for 6
years at 3½ % rate of interest.
5. Find the S.I. and the total amount for a principal of Rs. 10550 for 2
years at 10¼ % rate of interest.
6. Find the S.I. and the total amount for a principa l of Rs. 4360 for 4
years at 7.5% rate of interest.
7. Find the rate of interest if a person invests Rs. 1000 for 3 years and
receives a S.I. of Rs. 150.
8. Find the rate of interest if a person invests Rs. 1200 for 2 years and
receives a S.I. of Rs. 168.
9. A person invests Rs. 4050 in a bank which pays 7% S.I. What is the
balance of amount of his savings after ( i) six months, ( ii) one year?
10. A person invests Rs. 5500 in a bank which pays 10.5% S.I. What is the
balance of amount of his savings after 5 years?
11. A person invests Rs. 3000 in a bank which offers 9% S.I. After how
many years will his balance of amount will be Rs. 3135?
12. Find the principal for which the SI for 4 years at 8% is 585 more than
the SI for 3½ years at 11%. munotes.in
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Business
Mathematics
10 13. Find the principal for which the SI for 5 years at 7% is 250 more than
the SI for 4 years at 10%.
14. Find the principal for which the SI for 8 years at 7.5% is 825 more
than the SI for 6½ years at 10.5%.
15. Find the principal for which the SI for 3 years at 6% is 230 less than
the SI for 3½ yea rs at 5%.
16. After what period of investment would a principal of Rs. 12,350
amount to Rs. 17,043 at 9.5% rate of interest?
17. A person lent Rs. 4000 to Mr. X and Rs. 6000 to Mr. Y for a period of
10 years and received total of Rs. 3500 as S.I. Find ( i) rate o f interest, ( ii)
S.I. from Mr. X, Mr. Y.
18. Miss PankajKansra lent Rs. 2560 to Mr. Abhishek and Rs. 3650 to Mr.
Ashwin at 6% rate of interest. After how many years should he receive a
total interest of Rs. 3726?
19. If the rate of S.I. on a certain principal is same as that of the period of
investment yields same interest as that of the principal, find the rate of
interest.
20. If the rate of S.I. on a certain principal is same as that of the period of
investment yields interest equal to one -ninth of the principal, find the rate
of interest.
21. Find the principal and rate of interest if a certain principal amounts to
Rs. 2250 in 3 years and to Rs. 3750 in 5 years.
22. Find the principal and rate of interest if a certain principal amounts to
Rs. 3340 in 4 years and to Rs . 4175 in 5 years.
23. If Rs. 6400 amount to Rs. 7552 in 3 years at a certain rate of interest,
what will Rs. 8600 amount to in 4 years at the same rate of interest?
24. If Rs. 2700 amount Rs. 3078 in 2 years at a certain rate of interest,
what will Rs. 7200 amo unt to in 4 years at the same rate on interest?
25. At what rate on interest will certain sum of money amount to three
times the principal in 20 years?
26. Mr. Chintan earns as interest Rs. 1020 after 3 years by lending Rs.
3000 to Mr. Bhavesh at a certain rate on interest and Rs. 2000 to Mr.
Pratik at a rate on interest 2% more than that of Mr. Bhavesh. Find the
rates on interest. munotes.in
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Simple and Compound
Interest
11 27. Miss Bhagyashree lends at a certain rate on interest Rs. 2400 to Mr.
Shardul and Rs. 4200 to MrVishwas at rate of interest 3% less th an that of
Mr. Shardul. If after 5 years she receives Rs. 1890 as total S.I. find the
rates of interest.
28. Mr. Chaitanya invested a certain principal for 3 years at 8% and
received an interest of Rs. 2640. Mr. Chihar also invested the same
amount for 6 year s at 6%. Find the principal of Mr. Chaitanya and the
interest received by Mr. Chihar after 6 years.
29. Mr. Ashfaque Khan invested some amount in a bank giving 8.5% rate
of interest for 5 years and some amount in another bank at 9% for 4 years.
Find the amoun ts invested in both the banks if his total investment was Rs.
75,000 and his total interest was Rs. 29,925.
30. Mrs. Prabhu lent a total of Rs. 48,000 to Mr. Diwakar at 9.5% for 5
years and to Mr. Ratnakar at 9% for 7 years. If she receives a total interest
of Rs. 25,590 find the amount she lent to both.
31. Compute the compound amount and compound interest of Rs. 5000 if
invested at 11% for 3 years and the interest compounded i) annually, ( ii)
semi annually, ( iii) quarterly and ( iv) monthly.
32. Compute the compou nd amount and compound interest of Rs. 1200 if
invested at 9% for 2 years and the interest compounded i) annually, ( ii)
semi annually, ( iii) quarterly and ( iv) monthly.
33. Miss Daizy invested Rs. 25,000 for 5 years at 7.5% with the interest
compounded semi -annually. Find the compound interest at the end of 5
years.
34. Mr. Dayanand borrowed a sum of Rs. 6500 from his friend at 9%
interest compounded quarterly. Find the interest he has to pay at the end of
4 years?
35. Mr. Deepak borrowed a sum of Rs. 8000 from his friend at 8% interest
compounded annually. Find the interest he has to pay at the end of 3
years?
36. Mr. Deshraj borrowed Rs. 1,25,000 for his business for 3 years at 25%
interest compounded half yearly. Find the compound amount and interest
after 3 years.
37. Find the principal which will amount to Rs. 13468.55 in 5 years at 6%
interest compounded quarterly. [ (1.015)20 = 1.346855]
38. Find the principal which will amount to Rs. 30626.075 in 3 years at
7% interest compounded yearly. munotes.in
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Business
Mathematics
12 39. Find after how many years will Rs. 4000 amount to Rs. 4494.40 at 6%
rate of interest compounded yearly.
40. Find after how many years Rs. 10,000 amount to Rs. 12,155 at 10%
rate of interest compounded half -yearly.
41. Mrs. ManishaLokhande deposited Rs. 20,000 in a bank for 5 years. If
she re ceived Rs.3112.50 as interest at the end of 2 years, find the rate of
interest p.a. compounded annually.
42. On a certain principal for 3 years the compound interest compounded
annually is Rs. 1103.375 while the simple interest is Rs. 1050, find the
principal and the rate of interest.
43. On a certain principal for 4 years the compound interest compounded
annually is Rs. 13923 while the simple interest is Rs. 12000, find the
principal and the rate of interest.
44. A bank X announces a super fixed deposit scheme for its customers
offering 10% interest compounded half yearly for 6 years. Another bank
offers 12% simple interest for the same period. Which bank’s scheme is
more beneficial for the customers?
45. ABC bank offers 9% interest compounded yearly while XYZ bank
offers 7% interest compounded quarterly. If Mr. Arunachalam wants to
invest Rs. 18000 for 5 years, which bank should he choose?
Multiple Choice Questions:
1. How long it will take a sum of money invested at 5% p.a. simple
interest to increase its value by 30 %?
a) 5 years b) 6 years c) 7 years d) 8 years
2. The difference between the compound and simple interests on an
amount P at
r % p.a. for one year is Rs. ________.
a)
b) 0 c) 100 d) r – 1
3. The compound interests earned for two consec utive years, calculated
annually, are Rs. 800 and Rs.864 respectively. Therefore, the rate of
interest is
a) 10% b) 8% c) 6% d) 9%
4. A sum of money doubles itself at compound interest in 15 years. It will
become 4 times in
a) 30 years b) 60 years c) 45 years d) 75 years munotes.in
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Simple and Compound
Interest
13 5. Rs 1,200 is lent out at 9% p.a. simple interest for 3 years than the
amount after 3 years is __.
a) Rs. 1, 416 b) Rs. 1614 c) Rs. 1,564 d) Rs. 1,524
6. The difference in simple interest at 13% and 12% p.a. of a sum in one
year is Rs. 110. Then the sum is
a) Rs.11,000 b) Rs.15,000 c) Rs.12,000 d) Rs.14,000
7. The sum of money which increases 1/10 of itself every year and
amounts to Rs. 450 in 5 years at Simple interest than the sum is ___.
a) Rs.400 b) Rs.250 c) Rs.200 d) Rs.300
8. At what rate percent per annum will a sum of Rs 2000 amount to Rs
2205 in 2 years, compounded annually?
a) 7% b) 7.5% c) 5% d) 9%
9. Simple interest on a certain sum for 4 years at 7% p.a. is more than
simple interest on the same sum for 2.5 years at t he same rate by Rs. 840
then the principal amount is
a) Rs. 10,000 b) Rs. 8,000 c) Rs. 12,000 d) Rs. 9,000
10. A sum of Rs. 1600 lent at simple interest at 12.5% per annum will
become double in
a) 7years b) 8years c) 7.5years d) 10years
1.7 LIST OF REFERE NCES
Business Mathematics by Qazi Zameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook o f Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
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14 2
ANNUITY
Unit Structure
2.0 Objective
2.1 Introduction
2.2 Annuity
2.3 Present value
2.4 Future value
2.4.1 Sinking Fund
2.4 Equated Monthly Installments (EMI)
2.5 Let us sum up
2.6 Unit end exercise
2.7 List of References
2.0 OBJECTIVE
After goi ng through this chapter you will able to know:
The meaning and different terminologies of annuity
Different types of annuity
Derivation of formulas for different annuities
Calculation of present value of an annuity.
Calculation of future value and sinking fund.
EMI and method of calculation of EMI base on different types of
interest.
2.1 INTRODUCTION
In our day -to-day life we observe lots of money transactions. In many
transactions payment is made in single transaction or in equal installments
over a certai n period of time. The amounts of these installments are
determined in such a way that they compensate for their waiting time. In
other cases, in order to meet future planned expenses, a regular saving
may be done. Many people have had the experience of mak ing a series of
fixed payments over a course of time - such as rent, premium or vehicle
payments - or obtaining a series of payments for a course of time, such as
the certificate of deposit (CD) or intere st from a bond or lending money.
These ongoing or recurring payments are technically called "annuities”. munotes.in
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Annuity
15 2.2 ANNUITY
A series of equal periodic payments is called annuity . The payments are of
equal size and at equal time interval .
The common examples of a nnuity are: monthly recurring deposit schemes,
premiums of insurance policies, loan installments etc.
Types of Annuities:
Types of Annuities - There are three types of annuities.
• Annuity Certain: An annuity certain is an investment that provides a
series o f payments for a set period of time to a person or to the person's
beneficiary. It is an investment in retirement income offered by insurance
companies. The annuity may also be taken as a lump sum.
• Contingent Annuity: Contingent annuity is a form of annuit y contract
that provides payments at the time when the named contingency occurs.
For instance, upon death of one spouse, the surviving spouse will begin to
receive monthly payments. In a contingent annuity policy the payment
will not be made to the annuita nt or the beneficiary until a certain stated
event occurs.
• Perpetual Annuity or Perpetuity : A perpetual annuity, also called a
perpetuity promises to pay a certain amount of money to its owner forever.
Classification of Annuities: Annuities, in this sense o f the word, are
divided into 2 basic types: ordinary annuities and annuities due.
1. Ordinary Annuities (Immediate annuity): An ordinary annuity
makes (or needs) payments at the termination of each period. For example,
bonds usually pay interest at the termina tion of every 6 months.
2. Annuities Due: With an annuity due, payments, on the contrary come
at the start of each time period. Rent, which landlords typically need at the
initiation of each month, is one of the common annuity examples.
Note:
1. We consider only uniform and certain annuities.
2. If the type of an annuity is not mentioned, we assume that the annuityis
immediate annuity.
3. If there is no mention of the type ofinterest, then it is assumed that the
interest is compounded per annum.
There are va rious ways to measure the annuity rate changes or the cost of
making such payments or what they're ultimately worth. However, it is
first better to know about calculating the present value of the annuity or
the future value of the annuity.
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Business
Mathematics
16 2.3 PRESENT V ALUE
The sum of all periodic payments of an annuity is called its present value .
In simple words, it is that sum which if paid now will give the same
amount which the periodic payments would have given at the end of the
decided period. It is the one time pa yment of an annuity.
The formula to find the present value ( PV) is as follows:
Let i= x 100r
p, then the above formula can be rewritten as follows:
Example 1: Find the present value of an ordinaryannuity of Rs.70,000 p.a.
for 4 y ears at
14% p.a. compounded annually.
Answer: Given that C = Rs. 70,000, n= 4 years, r= 14%p.a.
•
the present value of an annuity immediate is given by
•
•
•
•
•
•
Example 2: Mr. Ashok Rane borrowed Rs. 20,000 at 4% p.a. compounded
annually for 10 years. Find the periodic payment he has to make.
Answer: Given PV = Rs. 20,000; n = 10; p = 1 and r = 4 i = 0.04 Where
P: periodic equal
payment
r: rate of interest p.a. p: period of annuity PV = 111npPii PV =
11
1 x 100 x 100npP
r r
pp
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Annuity
17 Now to find the periodic payments P we us e the following formula:
PV =
11
1npP
i i 20000 = 10110.04 (1 0.04)P= 0.04Px 0.3244 P = 20000 x 0.040.3244= 2466.09
Thus, the periodic payments are of R s. 2466.09
2.4 FUTURE VALUE
The sum of all periodic payments along with the interest is called the
future value (accumulated amount ) of the annuity.
The formula to find the future value ( A) of an immediate annuity is as
follows:
Here, P: periodic equ al payment
r: rate of interest p.a.
p: period of annuity i.e. yearly, half yearly, quarterly or monthly
and i = x 100r
p
Example 3: Find the future value after 2 years of an immediate annuity of
Rs. 5000, the rate of interest being 6 % p.a.
Answer: Given n = 2, P = Rs. 5000, r = 6 and p = 1 i = 6100= 0.06
A = P11ni
i= 500021 0.06 10.06= 50001.1236 10.06 A = 5000 x 2. 06 = Rs. 10300
Example 4: Find the future value of an immediate annuity after 3 years
with the periodic payment of Rs. 1200 at 5% p.a. if the period of payments
is (i) yearly, (ii) half -yearly, (iii) quarterly and (iv) monthly. A = P11 x 100
x 100npr
p
r
p= P11npi
i
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Business
Mathematics
18 Answer : Given P = Rs. 1200, n = 3, r = 5
(i) period p = 1 then i = 5100= 0.05
A = P11ni
i= 120031 0.05 10.05= 120001.1576 10.05 A = 1200 x 3.1525 = Rs. 3783
(ii) period p = 2 then i = 52 x 100= 0.025
A = P211ni
i= 120061 0.025 10.025= 12001.1597 10.025 A = 1200 x 6.388 = Rs. 7665.60
(iii) period p = 4 then i = 54 x 100= 0.0125
A = P411ni
i= 1200 121 0.0125 10.0125= 12001.1608 10.0125 A = 1200 x 12.864 = Rs. 15436.80
(iv) period p = 12 then i = 512 x 100= 0.00417
A = P1211ni
i= 1200 361 0.00417 10.00417= 12001.16147 10.00417 A = 1200 x 38.722 = Rs. 46466.83
Example 5:Mr. Nagori invested certain principal for 3 years at 8% interest
compounded half yearly. If he received Rs.72957.5 at the end of 3rd year,
find the periodic payment he made. [assume (1.04)6 = 1.2653]
Answer: Given n = 3, r = 8, p = 2 i = 82 x 100= 0.04
Now, A = P11npi
i 72957.5 = P61 0.04 10.04= P1.263 10.04 72957.5 = P[6.6325] munotes.in
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Annuity
19 P = 72957.56.6325= 11000
Thus, the periodic payment is Rs. 11,000
Example 6: Find the number of years for which anannuity of Rs.500 is
paid at the end ofevery year, if the accumulated amountworks out to be
Rs. 1,655 when interest iscompounded annually at 10% p.a.
Answer: Given thatC = Rs. 500, A = Rs. 1,655 n= ?
r= 10% p.a.,
Accumulated value is given by
1,655
1.331 =
n = 3 year
2.4.1 SINKING FUND
The fund (money) which is kept aside to accumulate a certain sum in a
fixed period through periodic equal payments is called as sinking fund .
We can consider an example of a machine in a factory which need s to be
replaced after say 10 years. The amount for buying a new machine 10
years from now may be very large, so a proportionate amount is
accumulated every year so that it amounts to the required sum in 10 years.
This annual amount is called as sinking fu nd. Another common example is
of the maintenance tax collected by any Society from its members.
A sinking fund being same as an annuity, the formula to compute the
terms is same as that we have learnt in section 1.11.
Example 7:A company sets aside a sum of Rs. 15,000 annually to enable
it to pay off a debenture issue of Rs. 1,80,000 at the end of 10 years.
Assuming that the sum accumulates at 6% p.a., find the surplus after
paying off the debenture stock.
Answer: Given P = Rs. 15000, n = 10, r = 6 i = 0.06 munotes.in
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Business
Mathematics
20 A = P(1 ) 1ni
i= 15000 x 10(1 0.06) 10.06= 15000 x 1.7908 10.06 A = Rs. 1,97,712
Thus, the surplus amount after paying off the debentur e stock
= 197712 – 180000 = Rs. 17712.
Example 8:Shriniketan Co -op Hsg. Society has 8 members and collects
Rs. 2500 as maintenance charges from every member per year. The rate of
compound interest is 8% p.a. If after 4 years the society needs to do a
work worth Rs. 100000, are the annual charges enough to bear the cost?
Answer: Since we want to verify whether Rs. 2500 yearly charges are
enough or not we assume it to be P and find its value using the formula:
A = P11ni
i
Here A = Rs. 1 00000, n = 4, r = 8 i = 0.08 P = x (1 ) 1nAii= 4100000 x 0.08(1 0.08) 1= 22192
Thus, the annual payment of all the members i.e. 8 members should be Rs.
22192. the annual payment per member = 221928= Rs. 2774
This payment is less than Rs. 2500 which the society has decided to take
presently. Thus, the society should increase the annual sinking fund.
2.5 EQU ATED MONTHLY INSTALLMENTS (EMI)
Suppose a person takes a loan from a bank at a certain rate of interest for a
fixed period. The equal payments which the person has to make to the
bank per month are called as equated monthly installments in short EMI.
Thus, EMI is a kind of annuity with period o f payment being monthly and
the present value being the sum borrowed .
We will now study the method of finding the EMI using reducing balance
method .
Let us recall the formula of finding the present value of an annuity.
PV =
11
1npP
i i
The e qual periodic payment ( P) is our EMI which is denoted it by E. munotes.in
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Annuity
21 The present value ( PV) is same as the sum ( S) borrowed.
Also the period being monthly p = 12.
Substituting this in the above formula we have:
S =
1211
1nE
i i
Thus, if S is the sum borrowed for n years with rate of interest r % p.a.
then the EMI is calculated by the formula:
Example 9:Mr. Sudhir Joshi has taken a loan of Rs. 10,00,000 from a
bank for 10 years at 11% p.a. Find his EMI using reducing balance
method.
Answer: Given S = Rs. 1000000, n = 10, r = 11 i = 0.11
Now, E = 12 x
11(1 )nSii= 1201000000 x 0.1111(1 0.11) = 1100000.9999 E = Rs. 1,10,000
Example 10: Mr. PrabhakarNaik has borrowed a sum o f Rs. 60,000 from
a person at 6% p.a. and is due to return it back in 4 monthly installments.
Find the EMI he has to pay and also prepare the amortization table of
repayment.
Answer : Given S = Rs. 60,000; n = 4 months;
r = 6% p.a. = 612= 0.5% p.m i = 0.005
Now, E = x 11(1 )nSii= 460000 x 0.00511(1 0.005) = 3000.01975
Now, we will prepare the amortization table i.e. the table of repayment of
the sum borrowed using reducing balance method.
In the beginning of the 1st month the outstanding principal is the sum
borrowed i.e. Rs. 60000 and the EMI paid is Rs. 15187.97 E = 12 x
11(1 )nSii
E = Rs. 15,187.97 munotes.in
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22 The interest on the outstanding principal is 0.005 x 60000 = Rs. 300 … (1)
Thus, the principal repayment is 1518 7.97 – 300 = Rs. 14887.97 … (2)
The outstanding principal ( O/P) in the beginning of the 2nd month is now
60000 – 14887.97 = 45112.03.
Note:
(1) is called the interest part of the EMI and (2) is called as the
principal part of the EMI.
As the tenur e increases the interest part reduces and the principal part
increases.
This calculation can be tabulated as follows:
Month O/P EMI Interest Part Principal
Part
(a) (b) (c) = (a) x i (b) - (c)
1 60000 15187.97 300 14887.97
2 45112.03 15187.97 225.56 14962.45
3 30141.02 15187.97 150.75 15037.22
4 15111.80 15187.97 75.56 15112.41
In the beginning of the 4th month the outstanding principal is Rs. 15111.80
but the actual principal repayment in that month is Rs. 15112.41. This
difference is due to roun ding off the values to two decimals, which leads
the borrower to pay 61 paise more!!
Example 11:Mr. ShyamRane has borrowed a sum of Rs. 100000 from a
bank at 12% p.a. and is due to return it back in 5 monthly installments.
Find the EMI he has to pay and al so prepare the amortization table of
repayment.
Answer: Given S = Rs. 100000; n = 5 months;
r = 12% p.a. = 1212= 1% p.m i = 0.01
Now, E = x 11(1 )nSii= 5100000 x 0.0111(1 0.01) = 10000.0485343= 20603.98
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23 The amortization table is as follows:
Month O/P EMI Interest Part Principal
Part
(a) (b) (c) = (a) x i (b) - (c)
1 100000 20603.98 1000 19603.98
2 80396.02 20603.98 803.96 19800.02
3 60596 20603.98 605.96 19998.02
4 40597. 98 20603.98 405.98 20198
5 20399.98 20603.98 204 20399.98
Flat Interest Rate (A layman’s approach) : In this method the amount is
calculated using simple interest for the entire period. This method is
hardly using any mathematical concepts except the fo rmula for simple
interest, so it is not accurate.
Example 12: A person brought Rs.75,000 at 12% p.a. if he wishes to
return the sum within a one year, Find his EMI using Flat interest rate
method.
Answer: Given that P = Rs. 75,000, r = 12%, n = 1 ye ar.
Amount =P + S.I. = 75,000 + 9,000 = Rs. 84,000
.
2.6 LET US SUM UP
In this chapter we have learn:
Definition of annuity and basic terms related to annuity.
Different types of annuity.
Formula and method of calculation of Present value an d future value.
Different method to calculate EMI.
2.7 UNIT END EXERCISE
1. Find the future value of an immediate annuity of Rs. 1200 at 6% p.a.
compounded annually for 3 years.
2. Find the future value of an immediate annuity of Rs. 500 at 8% p.a.
compounded p.m. for 5 years.
3. Find the accumulated amount of an immediate annuity of Rs. 1000 at
9% p.a. compounded semi -annually for 4 years. munotes.in
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Business
Mathematics
24 4. Find the future value of an immediate annuity of Rs. 2800 paid at 10%
p.a. compounded quarterly for 2 years. Also find the interest earned on
the annuity.
5. Find the periodic payment to be made so that Rs. 25000 gets
accumulated at the end of 4 years at 6% p.a. compounded annually.
6. Find the periodic payment to be made so that Rs. 30,000 gets
accumulated at the end of 5 years at 8% p.a. compounded half yearly.
7. Find the rate of interest if a person depositing Rs. 1000 annually for 2
years receives Rs. 2070.
8. Find the rate of interest compounded p.a. if an immediate annuity of
Rs. 50,000 amounts to Rs. 1,03,000 in 2 years.
9. Find the rate of interest compounded p.a. if an immediate annuity of
Rs. 5000 amounts to Rs. 10400 in 2 years.
10. What is the value of the annuity at the end of 5 years, if Rs. 1000 p.m.
is deposited into an account earning interest 9% p.a. compounded
annually? What is the interest paid in this amount?
11. What is the value of the annuity at the end of 3 years, if Rs. 500 p.m. is
deposited into an account earning interest 6% p.a. compounded
annually? What is the interest paid in this amount?
12. Mr. AshishGokhale borr ows Rs. 5000 from a bank at 8% compound
interest. If he makes an annual payment of Rs. 1500 for 4 years, what
is his remaining loan amount after 4 years?
(Hint: find the amount using compound interest formula for 4 years and
then find the accumulated amou nt of annuity, the difference is the
remaining amount.)
13. Find the present value of an immediate annuity of Rs. 10,000 for 3
years at 6% p.a. compounded annually.
14. Find the present value of an immediate annuity of Rs. 100000 for 4
years at 8% p.a. compound ed half yearly.
15. Find the present value of an immediate annuity of Rs. 1600 for 2 years
at 7% p.a. compounded half yearly.
16. A loan is repaid fully with interest in 5 annual installments of Rs.
15,000 at 8% p.a. Find the present value of the loan.
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Annuity
25 17. Mr. Suma n borrows Rs. 50,000 from Mr. Juman and agreed to pay Rs.
14000 annually for 4 years at 10% p.a. Is this business profitable to
Mr. Juman?
(Hint: Find the PV of the annuity and compare with Rs. 50000)
18. Mr. Paradkar is interested in saving a certain sum w hich will amount
to Rs. 3,50,000 in 5 years. If the rate of interest is 12% p.a., how much
should he save yearly to achieve his target?
19. Mr. KedarPethkar invests Rs. 10000 per year for his daughter from her
first birthday onwards. If he receives an interes t of 8.5% p.a., what is
the amount accumulated when his daughter turns 18?
20. Dr. Prabhulkar, a dentist has started his own dispensary. He wants to
install a machine chair which costs Rs. 3,25,000. The machine chair is
also available on monthly rent of Rs. 9 000 at 9% p.a. for 3 years.
Should Dr. Prabhulkar buy it in cash or rent it?
21. A sum of Rs. 50,000 is required to buy a new machine in a factory.
What sinking fund should the factory accumulate at 8% p.a.
compounded annually if the machine is to be replaced after 5 years?
22. The present cost of a machine is Rs. 80,000. Find the sinking fund the
company has to generate so that it could buy a new machine after 10
years, whose value then would be 25% more than of today’s price. The
rate of compound interest being 12% p.a. compounded annually.
23. Regency Co -op. Hsg. Society which has 15 members require Rs.
4,60,000 at the end of 3 years from now for the society repairs. If the
rate of compound interest is 10% p.a., how much fund the society
should collect from every member to meet the necessary sum?
24. Mr. Lalwaney is of 40 years now and wants to create a fund of Rs.
15,00,000 when he is 60. What sum of money should he save annually
so that at 13% p.a. he would achieve his target?
25. If a society accumulates Rs. 1000 p.a. from its 200 members for 5
years and receives 12% interest then find the sum accumulated at the
end of the fifth year. If the society wants Rs. 13,00,000 for society
maintenance after 5 years, is the annual fund of Rs. 1000 per member
sufficient?
26. If a so ciety accumulates Rs. 800 p.a. from its 100 members for 3 years
and receives 9% interest then find the sum accumulated at the end of
the third year. If the society wants Rs. 2,50,000 for society
maintenance after 3 years, is the annual fund of Rs. 800 per member
sufficient? munotes.in
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Business
Mathematics
26 27. Mr. Kanishk wants clear his loan of Rs. 10,00,000 taken at 12% p.a. in
240 monthly installments. Find his EMI using reducing balance
method.
28. Using the reducing balance method find the EMI for the following:
29. Mr. ArvindKamble has borrowe d Rs. 30,000 from his friend for 2
years at 14% p.a. If he is to return this amount in 5 monthly
installments, find the installment amount, the interest paid and prepare
the amortization table for repayment.
30. Mrs. Chaphekar has taken a loan of Rs. 1,25,000 from a bank at 12%
p.a. If the loan has to be returned in 3 years, find the EMI, Mrs.
Chaphekar has to pay. Prepare the amortization table of repayment of
loan and find the interest she has to pay.
31. A loan of Rs. 75,000 is to be returned with interest in 4 installments at
15% p.a. Find the value of the installements.
32. Find the sum accumulated by paying an EMI of Rs. 11,800 for 2 years
at 10% p.a.
33. Find the sum accumulated by paying an EMI of Rs. 1,800 for 2 years
at 12% p.a.
Find the sum accumulated by p aying an EMI of Rs. 12,000 for 3 years
at 9% p.a.
Multiple Choice Questions:
1. The accumulated value of an annuity immediate , of Rs.20,000 p.a. for
3 years at 10% p.a. compounded yearly, is Rs.
a) 66,200 b) 66,221 c) 92,820 d) 60,000
2. An annuity i n which each payment is made at the end of the year is
called ________.
a) annuity due b) annuity certain
c) Immediate annuity d) uniform annuity
3. A loan of Rs.80,000 is returned in 3 monthly installments at 12% p.a.
find the EMI using the flat ra te.
a) Rs.25,488.67 b) Rs.27,466.67 c) Rs.28,576.67 d) Rs. 26,567.82
4. The present value is always _______ the future value.
a) More t hen b) less then c) equal to d) independent of
5. An annuity in which the number of payments is fixed is called
a) Fix annuity b) limited annuity
c) certain annuity d) immediate annuity. munotes.in
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Annuity
27 6. An annuity in which all the payments are equal is called
a) Equitable annuity. b) egalitarian annuity.
c) annuity due. d) uniform annuity.
7. The presen t value of an immediate annuity of Rs.50,000 p.a. for 3 years
at 10% p.a. compounded annually is Rs.
a) 1,80,000 b) 1,24,342.60 c) 1,24,234.06 d) 1,50,000
8. The accumulated amount after 3 years of an immediate annuity of
Rs.5,000 p.a. with the rate of interest of 6% compounded annually is
Rs.
a) 15,000 b) 15,900 c) 15.921.23 d) 15,918
9. If the payment of annuity are made at the beginning of each period, the
annuity is called
a) Ordinary annuity b) Annuity due
c) Uniform annuity d) Immediate an nuity
10. When the EMI are calculated using present value of the annuity using
compound interest, the method is called
a) Flat rate method b) Repayment method
c) Reducing balance method d) Amortization method
2.8 LIST OF REFERENCES
Busines s Mathematics by QaziZameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
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28 3
FUNCTIONS
Unit Structure
3.0 Objective
3.1 Introduction
3.2 Function
3.3 Types of function
3.4 The Function used in Business and Economics.
3.5 Break -even point
3.6 Equilibrium point
3.7 Let us sum up
3.8 Unit end exercise
3.9 List of Reference s
3.0 OBJE CTIVE
After going through this chapter you will able to know:
Definition of function and types of functions.
Different types of function used in business and economics.
Most important concept of business break -even point.
The concept of equilibri um point.
3.1 INTRODUCTION
A function was the heart of the scientific revolution of the seventeenth
century. To understand the general use of function we must study their
properties in the general, which is what we do in this chapter.
The reader is no doub t familiar with function of the form ()yf x for
instance, if 2() 2fx x, 2x and (2)yf then the value of y is 4.
3.2 FUNCTION
A function can be intuitively understood as a relation between two
variables, where one variable which depends on the other is called as
dependent variable while the other is called as independent variable. Let y
be variable depending on an independent variable say x, then the variable
y is said be a function of x which can be denoted as y = f(x) (This is
pronounced as y equal to f of x). munotes.in
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Functions
29
We all know that demand of a certain product depends on its price and
vice-versa. If D denotes demand of a certain product and P denotes its
price then the relation between demand and price can be written as D =
f(P) or P= g (D). Here f and g are notations for showing the function
(relation).
For e.g., for a certain commodity the relation between demand and price is
given askDP, which means that , as the price of the commodity increases
its demand falls and as the price decreases the demand increases.
3.3 TYPES OF FUNCTION
Constant function:
A function which takes a constant value is called as constant function .
Example: f(x) = 4, g(x) = a, where a is any real value.
Graphically , a constant function is a straight line parallel to the X -axis or
the Y -axis. In general, for a fixed real value k, a constant function is
written as f(x) = k.
In both the examples we can see that the function takes a fixed value over
all values of the independent variable x.
The graph of f(x) = 4 is as follows:
Linear function
A function whose graph is a straight line is called as a linear function .
In general, for fixed real values m and c, the function f(x) = mx + c is a
linear function. Here m is the slope of the line and c is its y -intercept.
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30
Example: f(x) = 2 x + 3 . The graph of this function is as shown below:
Step function
A function which takes different constant values at different intervals of
the independent variable is called a step function . The name step function
is being given because of the graphical behavior of the function.
Example: ( i) Consider the
function ( ) 1 for 0 1fx x 2 for 1 2x 3 for 2 3x
ii) Consider the function
() 3gx 01x 2 12x 1 23x
Power function
A function which has the dependent variable as a power of the
independent variable is called as a power function . Here the variable ( x) is
the base and a natural number ( n) as the index (power).
Example: 2()fx x, 3() 3gx x munotes.in
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Functions
31
Exponent ial function
A function which has a constant base and variable index is called as an
exponential function . If a is any positive integer then ()xfx ais an
exponential function .
A very common example in calculus is when the number e whose
approximate value is 2.718 is taken as the base. i.e. ()xfx e.
The graph of xe is given below:
Logarithmic function
Logarithmic function is an inverse function of exponential function.
If yxawhere a> 0 and1a, then logayxis the logarithmic
function. Here y is said to be the logarithm of x to the base a.
Ifyxe, then log logeyx x is the logarithmic function. Here y is
said to be the logarithm of x to the base e. Usually, when e is taken as the
base it is not written.
If m and n are positive numbers, then
1. log ( ) log logaa amn m n
2. log log logaa ammnn
3. log ( ) logaanmnm
4. log 1 0a
5. log 1aa munotes.in
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Mathematics
32 6. logloglogabaxxb
Example .1: If 2() 1fx x x , find the value of f at x = -1, 0, 1, 2
Answer : At x = -1, 2(1 ) (1 ) (1 ) 1f 1111
(1 ) 1f
At x = 0, 2(0) 0 0 1 1f (0) 1f
At x = 1, 2(1) 1 1 1 3f (1) 3f
At x = 2, 2(2) 2 2 1 7f (2) 7f
Example .2. If 2() 2fx x, find f(x + 1) and f(x – 1)
Answer : Since2() 2fx x
2(1 ) (1 )2fx x
22(1 ) 21 2 23fx x x x x
Similarly,
2(1 ) (1 )2fx x
22(1 ) 21 2 23fx x x x x
Example .3. Find the value of x, if f(3x) = 3 f(x) where
2() 3fx x x .
Answer : Given 2() 3fx x x 23( ) 3 3 9fx x x
Also, 22(3 ) (3 ) (3 ) 3 9 3 3fx x x x x
Also given that, f(3x) = 3 f(x)
2293 3 33 9xx xx
266 0x 210x
21x
1x
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Functions
33 Example 4. If f(x) = log x, show that f(xy) = f(x) + f(y)
Answer : ( ) logfx xQ
( ) logfy y
Also, ( ) log( )f xy xy … (1)
Using the properties of logarithms, we have
log( ) log logxy x y … (2)
Thus, ( ) log logfx y x y … from (1) and (2)
() ( ) ( )fx y fx fy . Hence proved.
3.4 THE FUNCTION USED IN BUSINESS AND
ECONOMICS
Demand function
Let p be the price of a certain commodity and Dbe its demand. As per the
economic co nvention price is considered as a function of demand and
hence we can denote this relation as p = f(D) or p = f(Qd). (Qd is the
quantity of demand). This is called as the demand function .
The graph of demand function called as demand curve, is drawn by tak ing
the price ( p) on the Y -axis and the demand ( D) on the X -axis.
The graph of demand function always has a negative slope, i.e. demand
curve falls downwards from left to right indicating that as the demand
increases the price of that commodity decreases.
Examples:37pD, 5pD, 23 4 11pD D , 242pD or 53dQp, 101dQp, 224dQp p etc.
Supply function
If p is the price of a certai n commodity and S is its supply, then the supply
function is given by p = f(S) or p = f(QS), where QS is the quantity of
supply.
The graph of supply function called as supply curve, is drawn by taking
the price( p) on the Y -axis and the supply ( S) on the X -axis.
The graph of the supply curve has a positive slope, i.e. the supply curve
move upwards from left to right.
Examples:10 4pS, 234pS , or 23SQp, 23SQp petc. munotes.in
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34 Total Cost func tion
If x is the number of units produced of a certain commodity and C is the
total cost incurred then the cost function is given as ()Cf x
Thus, cost depends on the quantity produced.
The cost is divided into two parts:
a) Fixed cost ( FC): This is independent of the quantity produced. e.g.
labor cost, maintenance, rent etc.
b) Variable cost ( V(x)): This depends on the quantity produced.
Thus, C = FC + V(x).
Examples: 76Cx, 225Cx x.
Average Cost functi on
If x is the number of units produced of a certain commodity and C is the
total cost, then the cost per unit of the commodity is called as the average
cost (AC).
Average cost = ()CxACx
Total Revenue
If p is the price per unit and D is the demand of a certain product, then the
total revenue function (TR) is given by TR pD
Average Revenue
Average revenue ( AR) is the revenue per unit demand of the product. In
other words, AR is the price per unit of the product.
Rp DAR pDD AR p
Example 5. The demand function for a product is given by 264pD,
where p is the price per unit of the product and D is the demand. Find the
total revenue, average revenue for the product.
Ans: Let R be the total revenue function.
w.k.t. R = pD
2(6 4 )RD D
364RDD
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Functions
35 And average revenue is given by, AR = p
264AR D
Example .6. If the demand function of a product is given by
26 11pD D, find the total revenue when D = 5.
Ans: Given 26 11pD D
2(6 1 1 )R pD D D D
3261 1RD D D
When D = 5,
32
55 6(5) 11(5)DR
5125 150 55 330DR .
3.5 BREAK -EVEN POINT
Profit function
If R is the total revenue and C is the total cost of a product, then the profit
function denote as , is given by π = R – C.
If π> 0 then there is a profit.
If π < 0 then there is a loss.
If π = 0 then there is a situation wherein ther e is no loss or no profit.
Such a point is called as
Break Even Point (BEP). In other words, at the BEP the total revenue is
same as the total cost.
Now we know that, π = R – C.
If π = 0, R – C = 0 RC.
Example 7. The total cos t C( in lakhs of Rs.) in manufacturing x units of
a cell phone is given by 0.055 219Cx. Each cell phone is sold at price
of Rs. 8,500. Find the profit function and the break even point .(BEP)
Ans: The revenue function is given by 8500 x 100000xR
0.085Rx
The profit function is given by π = R – C
0.085 (0.055 219)xx
0.085 0.055 219 0.03 219xx x munotes.in
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36 0.03 219x .
The BEP is achieved when π = 0
Thus, 0.03 219 0x
0.03 219x
7300x
Hence the BEP is achieved when 7,300 cell phones are manufactured.
Example 8. The demand function for a product is given bypa xb. If
the prices for producing 1unit and 6 units of the product are 10 and 30
respectively, find the demand fu nction explicitly. Also find the total
revenue when x = 10.
Ans: When x =1, the demand function is 10 = a + b … (1)
When x = 6, the demand function is 30 = 6 a + b … (2)
Solving the eqns (1) and (2), we get
a = 4 and b = 6.
Hence the de mand function is p = 4x + 6
The total revenue is 2(4 6) 4 6R px x x x x
When x = 10, 2
104(10) 6(10) 400 60 460xR
Example 9 The fixed cost for manufacturing a product is Rs. 45,000 and
the variable cost is Rs. 500 per unit of the product. The revenue fun ction is
given by 2100 3600Rx x . Find the profit function, BEP and the values of
x for which there will be profit.
Ans: The total cost function is given by:
()C FC V x
From the given data we have,
45000 500Cx … (1)
The revenue function is given as:
2100 3600Rx x … (2)
Thus, the profit function π = R – C from (1) and (2) is,
2100 3600 (45000 500 )xx x
2100 4100 45000xx … (3) munotes.in
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Functions
37 The BEP is achieved when0. Equating eqn (3) to 0, we get
2100 5900 45000 0xx
Dividing by 100 on both sides, we get
241 450 0xx
( 9)( 50) 0xx … (4)
x = – 9 orx = 50
Discarding x = – 9 , as number of units produced cannot be negative, t he
BEP is at x = 50.
We know that, the manufacturer will earn profit when the sales are above
the BEP values.
Thus, profit is achieved for any value of x greater than 50.
3.6 EQUILIBRIUM POINT
As we have seen above that price is a function of demand as we ll as
supply. If both the curves are drawn for the same commodity, we see that
the curves intersect at a point. This point, where the supply and demand is
equal is called as the equilibrium price denoted as ep.
Example 10. The deman d function for a certain product is given by 72pD and the supply function is given by 45pS, where p is the
price per unit, D is the demand and S is the supply. Find the equilibrium
price and quantity for the prod uct.
Ans:We will first write the demand and supply functions as functions of
price.
72pD 72pD …(1)
And 45pS 54pS …(2)
Now, w.k.t., equilibrium price is that pri ce where the demand and supply
coincides. So we equate both the above equations,
i.e.7524pp
4(7 ) 2( 5)pp
28 4 2 10pp
6 18p 3p munotes.in
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38 Thus the equilibrium price is pe = 3.
To find the equilibrium quantity, we substitute the equilibrium price in any
of the eqns (1) and (2);
Substituting p = 3 in eqn (1), we get
77 3124DpQD .
Thus the equilibrium quantity is 1DQ.
3.7 LET US SUM UP
In this chapter we have learn:
Function and its types.
Different business and economics functions
To solve different Problem to using function.
The Concept of Break -even point and equilibrium point.
3.8 UNIT END EXERCISE
1. Define dependent variable and i ndependent variable.
2. Explain the demand and supply function with proper examples from
economics.
3. Write a short note on Break even Point .
4. Classify the following functions as constant function, linear function,
power function, exponential function, step function and logarithmic
function:
(i) ( ) log10fx, (ii) ( ) 10 15fx x(iii) () 3xfx(iv) ()xfx e
(v) ( ) 4 0 2 = 8 2 4 = 12 4 6fx x
xx
5. Given that 2() 2 3fx x x , find f(x + 2).
6. If 2() 3 6fx x, show that (2 ) 4 ( ) 18 0fx f x .
7. Find x, if(2 ) 4 ( )fx f x, where2() 3 6fx x x .
8. If 23()1xfxx, find 231xfx.
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Functions
39 9. If 1() l o g1xfxx, show that () ()1xyfx fy fxy
10. If 1() l o g1xfxx, show that () ()1xyfx fy fxy
11. If 1() l o g1xfxx, show that () ()1xyfx fy fxy
12. If the demand function is 15DQp and the supply function is
103SpQ, find the equilibrium price and equilibrium quantity.
13. If the demand function is 21 2DQp and the supply function is 31 8SQp, find the equilibrium price and equilibrium quantity.
14. If the deman d function is 12 110DQp and the supply function is
152SpQ, find the equilibrium price and equilibrium quantity.
15. Find the equilibrium price if 82DpQpand2SQp.
16. If the demand funct ion is 10pD and the supply function
is20 2pS, find the equilibrium price and equilibrium quantity.
17. If the demand function is 3 16pD and the supply function
is 20pS, find th e equilibrium price and equilibrium quantity.
18. The demand functions for two products A and B are as follows:
10ABDpand 12BA BDp p . The corresponding supply functions are
2AA BSp p and 2BA BSp p . Find the equilibrium prices and
equilibrium quantities.
19. Mr. BhaveshShah who owns a Tea factory, finds that variable cost
for processing of1 gm of tea is Re.1 and his fixed costs are Rs. 3000.
Write down the cost function, hence find the cost for processing 1kg of
Tea.
20. Mr. Gunderia’s glass factory has 35 workers. The cost of producing 1
sheet of glass is Rs. 34.50. The fixed cost per worker is Rs. 25. If the glass
sheet is sold at Rs.52, find the number of sheets to be produced so that
there is no loss. If a discount of Rs.2 is given per sheet what is the BEP?
At this price what is the profit if 200 sheets are produced?
21. A firm has the cost function 2( 2)Cx xand the revenue function 214Rx x. Find the average cost, average revenue and profit function.
Hence find the profit at the BEP.
munotes.in
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40 22. The price of a chocolate was Rs. 5 when its demand was 10,000. Its
demand came down to 8,000 at the price of Rs.8 per chocolate. If the price
is a linear function of t he demand, find the explicit relation. Estimate the
demand when the price will be Rs. 10 per chocolate. Which of these three
prices gives more profit for the manufacturer?
23. The total cost and the total revenue of a firm is given by 53 5 0Cxand250Rx x. Find the break even points.
24. Mr. VinodMenon manufactures wall clocks. The cost of setting up the
factory is Rs. 5 lakhs. The cost of production per wall clock is Rs. 100.
Determine the cost function. If each wall clock is sold at Rs. 250, write
down the revenue and profit function. If 2000 walls clocks are produced
and sold does Mr. Vinod earn any profit? If so, what is his profit? Also
find the BEP.
25. Mr. Roy manufactures pens, whose total cost is given by 42 5 2 0 0Cx. If each pen is sold at Rs. 12, find the minimum number of
pens Mr. Roy should produce so that there is no loss. If a discount of Rs. 2
is given per pen, what would be the BEP? If 5000 pens are sold everyday
at the discounted cost, does Mr. Roy earn any profit? If yes, find the profit.
Multiple Choice Questions:
i) For the demand function D=20 -5p+p2 then the demand when p=3 is___
a) 16 b) 15 c) 14 d)10
ii) The cost function is C=x2+x+100 then the average cost of p roducing 10
items is ___
a) 20 b) 21 c) 210 d) 200
iii) If f(x)=3x+5 \x then the value of f(x) at x=0 is______
a) 0 b) 1 c) 3 d) function does not exist
iv) If the total cost function is C=4x2+x+8 then the average cost when x=2
is____
a) 13.75 b) 13 c) 26 d) 24
v) For the function
is
a) constant function b) linear function
c) quadratic function d) identity f unction
3.9 LIST OF REFERENCES
Business Mathematics by Qazi Zameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
munotes.in
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41 4
PERMUTATION AND COMBINATION
Unit Structure
4.0 Objective
4.1 Introduction
4.2 Prerequisites and terminology
4.3 Permutation
4.4 Formula to compute permutations of ndifferent objects
4.5 Permutations of n objects not all different
4.6 Combination
4.7 Let us sum up
4.8 Unit end exercise
4.9 List of References
4.0 OBJECTIVE
After going through this chapter you will able to know:
Factorial notation.
Definition of permutation and its formula.
Arrangements of n different objects may be all are same or distinct.
Definition of Combination and its formula.
Using combination we can calculate different ways of selection.
4.1 INTRODUCTION
We often hear people saying that ‘probably it will rain today’; ‘it is likely
that India will win the match against Au stralia on this pitch’; ‘the chance
of passing the CET is 10% only’ and so on. The words ‘probably’, ‘by
chance’ or ‘likely’ are statistical terms but are very commonly used by all
of us. It is very natural that people are interested to know about the
possibility that something happens. People interested in sports are eager to
know, the possibility of their team to win a game; political activists want
to be sure their chances of winning an election, meteorological department
would like to know about the wea ther, an economists may want to know
the chance that sales will increase if the price of a commodity is decreased
etc. Also it becomes necessary sometimes to know in how many different
ways a particular event may happen. munotes.in
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42 All these calculations can be broa dly classified into two types namely
permutations and combinations .
4.2 PREREQUISITES AND TERMINOLOGY
Before going into the detailed study of these two methods, we shall
discuss some of the prerequisites which are useful in understanding the
concepts and f ormulae related to them.
Factorial:
The factorial of a natural number n is defined as the product of all
numbers from 1to n. It is denoted as n!
For example :
Factorial of 3 i.e. 3! = 3 x 2 x 1 = 6 ( This is read as 3 factorial equal to
6)
Similarly, 5! = 5 x 4 x 3 x 2 x 1 = 120
In general, n! = n(n – 1)(n – 2)(n – 3)……3 x 2 x 1
This formula gives a recursive relation: n! = n(n – 1)!
For example :
6! = 6 x 5 x 4 x3 x 2 x 1 = 6 x 5!
Remark : We define 0! = 1
Fundamental Principal of Counting :
If there a re m ways doing one thing and n ways of doing another thing
then the total number of ways of doing both the things together is mn.
4.3 PERMUTATION
A permutation of n objects is an arrangement of some of these (or all)
objects in a definite order. The order in which the arrangement is done is
important in permutations.
Example 1: In how many different ways can three friends Mitesh, Ritesh
and Paresh stand for a group photograph?
Ans: Let us denote these friends by their first alphabet M, R and P. The
three f riends can be arranged as shown below:
M R P M P R R M P R P M P M R P RM
The number of ways is 6.
One should not be satisfied with this answer. The question that should
come to our mind is how did we arrange them so? Well, if we observe the
arrangement ag ain, it can be seen that first M’s place was fixed and the munotes.in
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Permuta tion and
Combination
43 remaining two were arranged. This step was repeated again for R and P. In
terms of permutation what we did was arranging 3 objects amongst
themselves. The next question should be what if there are 10 friends? Can
you write down their different arrangements explicitly as above?
4.4 FORMULA TO COMPUTE PERMUTATIONS OF
NDIFFERENT OBJECTS
nP r: The number of ways of arranging r objects out of n objects is denoted
by n P r and is calculated by the formula : n P r = !() !nnr
For e.g: 4
24! 4 x 3 x 2 x 112(4 2)! 2 x 1P
Remark : The number of ways of arranging all n objects is thus !!() ! 0 !n
nnnPnn= n!
Now if we go back to the first example, where we had to arrange 3 friends,
then the number of ways of their arrangement using the above formula is 3! 6, which is the same answer what we had got by arranging them
explicitly.
Example 2: In how many ways can 6 people be photographed, if only 4
can be seated at a time?
Ans: Here n = 6 and r = 4. no. of ways of arranging 4 out of 6 people = 6
46!(6 4)!P= 6 x 5 x 4 x 3 x 2 x 12 x 1= 720
Example 3: If 3542nnPP then find the value of n.
Ans:5!5!n nPn and 3!3!n nPn
Since 3542nnPP !!425! 3!nn
nn 3!425!n
n
i.e.3( 4 ) ( 5 ) !425!nnn
n (n – 3)(n – 4) = 42 munotes.in
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44 n2 – 7n + 12 = 42 n2 – 7n – 30 = 0 (n – 10)(n + 3) = 0 n = 10, – 3
As n ≠ – 3 n = 10
Example 4: Show that 11.nn
rrPn P
Ans: Consider RH.S. = 11.nrnP = n. (1 ) ![1 ) ( 1 ] !n
nr
= (1 ) !11 !nn
nr= !!nnr= nrP= L.H.S.
Example 5: In how many different ways can a 4 digit number be formed
from the numbers 1, 2, 3, …, 9, with no digit being repeated?
Ans: Since no repetition is allowed, the number of ways of forming a four
digit number from the given 9 digits is
94P= 9!94 != 9 x 8 x 7 x 6 x 5!5!= 302 4.
Example 6: In how many different ways can the letters of the word
“MATHS” be arranged if no letter is to be repeated in the same word?
Ans: The word “MATHS” consists of four letter M, A, T, H and S.
The required number of ways is arranging 4 objects all at a time.
Thus, the no. of different ways = 4! = 24.
Example 7: Eight candidates are to appear for an interview in a company.
In how many ways can the HR manager schedule the candidates for their
interview? What if two of the eight candidates are not be interviewed?
Ans: The number of ways of arranging 8 candidates is 8! = 40320
If two out the eight are not be arranged, it means to arrange the remaining
six out of 8.
This can be done in 86Pways. the number of ways of scheduling the candidates now = 8!86 !
= 8!2!= 403202= 20160 munotes.in
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Permuta tion and
Combination
45 Example 8: Find the number of ways in which 3 books on Economics, 4
books on Mathematics and 2 books on Law can be arra nged on a book
shelf so that books of the same subject are together. Find the number of
arrangements if books are to be arranged at random.
Ans: We can consider this problem of arranging three blocks E (books on
Economics), M (books on Mathematics) and L (books on Law), which can
be done in 3! ways.
Now,
3 books in Economics can be arranged amongst themselves in 3! ways,
4 books in Mathematics can be arranged amongst themselves in 4! ways,
and 2 books in Law can be arranged amongst themselves in 2! Ways
Thus, by the fundamental principle of counting,
the total number of arrangements = 3! x 3! x 4! x 2! = 1728 ways.
If there is no condition on the position of any book of any subject we may
consider this of arranging 3 + 4 + 2 = 9 books, which can be done in 9!
Ways
i.e. 362880 ways.
Example 9: In how many ways can 3 boys and 4 girls be seated for a
group photograph if (i) no two boys sit together, (ii) no two girls sit
together, (iii)all boys sit together
Ans: (i) no two boys sit together
4 girls can be arran ged in 4! = 24 number of ways
In each such arrangement there are five places (marked as X) where boys
can be seated so that no two boys sit together as shown below:
X G X G X G X G X
Now, 3 boys can be seated in these 5 places in 53Pways = 60 ways
Thus, the total number of arrangements = 24 x 60 = 1440.
(ii) no two girls sit together :
3 boys can be arranged in 3!number of ways
In each such arrangement there are four places (marked as X) where girls
can be seated so that no two girls s it together as shown below:
X B X B X B X
Now, 4 girls can be seated in these 4 places in 44Pways = 4! ways munotes.in
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46 Thus, the total number of arrangements = 3! x 4! =144.
(iii) all boys sit together
All boys can be considered as one block an d the remaining 4 girls as four
blocks. The required arrangement is of these 1 + 4 = 5 blocks, which can
be done in 5! ways .
The 3 boys can be arranged amongst themselves in 3! ways .
Thus the total number of ways in which 3 boys and 4 girls can be seated
for a group photograph such that all boys are seated together is = 5! x 3! =
720.
Example 10: In how many different ways can the letters of the word
“JOGESHWARI” be arranged such that (i) there is no restriction, (ii) the
word starts with ‘A’, (iii) the wor d ends with ‘W’, (iv) the word begins
with ‘A’ and ends with ‘W’, (v) the vowels are together, (vi) the letters W,
A, R are never together.
Ans: The given word “JOGESHWARI” consists of 10 distinct letters, of
which 4 are vowels (A, E, I, O) and 6 are conso nants ( G, H, J, R, S, W)
(i) If no restriction is there then the total number of arrangements = 8! =
40320
(ii) The word starts with ‘A’
Since the first place is fixed for the 10 letter word, it remains to
arrange the remaining 9 letters.
Thus, the total numbe r of ways = 9! = 362880
(iii) The word ends with ‘W’
This is similar to the above problem
(iv) the word begins with ‘A’ and ends with ‘W’
Since two letters A and W are fixed, it remains to arrange the remaining 8
letters which can be done in 8! = 40320 wa ys.
(v) the vowels are together
4 vowels can be considered as one block and the remaining 6 consonants
as 6 blocks.
Thus, the problem now reduces to arranging these 7 blocks, which can be
done in 7! = 5040 ways.
The 4 vowels can be arranged amongst themsel ves in 4! = 24 ways.
Thus by fundamental principle of counting, the total number of ways =
5040 x 24 = 120960.
(vi) the letters W, A, R are never together munotes.in
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Permuta tion and
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47 The required number of ways = total number of all arrangements – number
of ways of arrangements wher e W, A, R are together.
Now, the total number of ways of arranging 10 letters = 10!
No of ways of arranging the letters W, A, R together = 8! x 3! (from the
above case (v))
Thus the required number of ways = 10! – (8! x 3!)= 3628800 – 241920
= 33868 80.
4.5 PERMUTATIONS OF N OBJECTS NOT ALL
DIFFERENT
Consider a permutation of 2 white and 4 black hats of the same type. Then
the permutation W 1W2B1B2B3B4 is same as W 2W1B1B2B3B4 as we cannot
differentiate between the two white hats. Thus, in permutations where
certain numbers of objects are similar, we have to remove the duplications
of same permutations.
Let n1 objects be of one kind, n2 objects be of second kind,…, nk objects
be of kth kind then the number of distinct permutations of all objects taken
together i.e. n = n1 + n2 + … + nk is given by:
123!! ! !...... !knnnn n
Example 11: A College Librarian Mrs. Akalpita orders 15 books of
which 5 books are of Business Law, 5 books of Business Mathematics, 3
books on Principles of Management and 2 books o n F.H.S. In how many
different ways can she arrange them on the shelf?
Ans: Given n = 15, n1 = 5, n2 = 5, n3 = 3 and n4 = 2
The number of ways of arranging the 20 books on the shelf
= 15!5!5!3!2!= 7567560
Example 12: In how many different w ays can the letters of the word
“MALAYALAM” be arranged horizontally?
Ans: In the given word “MALAYALAM” has n = 9 letters of which M, A
and L are repeating.
Here ‘M’ is repeated 2 times, n1 = 2
‘A’ is repeated 4 times, n2 = 4
‘L’ is repeated 2 times, n3 = 2 the number of permutations = 9!2!4!2!= 3780. munotes.in
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48 Example 13: Find the number of permutations of the letters of the word
“TOMMORROW” such that (i) no two M’s are together, (ii) all the O’ s
are not together.
Ans: The given word “TOMMORROW” has 9 letters of which ‘O’, ‘M’
and ‘R’ are repeated.
Now, ‘O’ is repeated 3 times, n1 = 3
‘M’ is repeated 2 times, n2 = 2
‘R’ is repeated 2 times, n3 = 2 number of permutations of letters of the word “TOMMORROW”
= 9!3!2!2!= 15120 … (1)
(i) no two M ’s are together
We assume the two M’s as one block and the remaining 7 letters as
remaining 7 blocks.
The number of ways of arranging these 7+1 = 8 blocks is 8!, in which the
letters R and O are repeated 2 and 3 times respectively.
Hence, the number of permutations in which both the M’s are together is
=8!3!2!= 3360. … (2)
From (1) and (2), the number of distinct permutations in which no two M’s
are together = 15120 – 3360 = 11760.
(ii) all the O’s are not together
We consider the three O’s as one block and the remaining 6 letters as one
block.
The number of ways of arranging the 6 + 1 = 7 blocks is 7!, in which the
letters M and R are repeated 2 times each.
Hence, the number of permutations in which all the O’s are together is
= 7!2!2!= 1260 … (3)
From (1) and (3), the number of distinct permutations in which all three
O’s are not together = 15120 – 1260 = 13860
Example 14: Find the number of permutations of the letters of the word
“VOWEL” such that the vowels occupy the odd places in the arrangement.
Ans: The given word “VOWEL” has 2 vowels ‘E’, ‘O’ and 3 consonants.
There are two odd places viz. 1st, 3rd and 4th. munotes.in
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Permuta tion and
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49 1st ___ 3rd ____ 5th
Two vowels can be placed in three places in 32P ways. For each such way
the 3 consonants can be placed in the remaining three places in 3! ways.
By Fundamental principal of coun ting, the total number of arrangements is
given by 32Px 3! = 6 x 6 = 36 ways.
4.6 COMBINATION
A combination of n objects is an arrangement of some of these (or all)
objects where the order of arrangement is not considered.
For e.g.: A combination considers the arrangements ‘ ab’ and ‘ ba’, of two
letters ‘ a’ and ‘ b’ as the same.
nrC: The selection (or rejection) of r objects out of n objects is denoted
by nrCand is calculated by the formula: nrC= !!( )!nrn r
For example: 52C= 5!2!(5 2)!= 5!2!3!= 5 x 4 x 3 x 2 x 12 x 3 x 2 x 1= 10
We know that the factorial formula can be written recursively. Hence
we can write the numerator n! asn(n – 1)(n – 2)….( n – r + 1)( n – r)! Due
to this recursive formula, the calculation of nrCbecomes easier as show
below:
Now, nrC= ( 1)( 2)....( 1)( )!!( )!nn n n r n rrn r
= ( 1)( 2)....( 1)!nn n n r
r
For example:104C= 10 x 9 x 8 x 7 4 x 3 x 2 x 1= 210. This method simplifies and
speeds up the calculations
Results (without proof) :
1. 0nC= nnC= 1
2. nnrn rCC
3. 1nn
rrCC= 1nrC
Example 15: In how many ways can 2 students be selected for a student’s
committee out of 7 students?
Ans: Here n= 7 and r = 2 the no. of ways of selecting 5 out of 7 students = 72C= 7 x 62 x 1= 21. munotes.in
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50 Example 16: A question paper has 8 questions and only 5 questions are to
be attempted. I how many ways a student can select any 5 questions?
Ans: Here n = 8, r = 5 the required nu mber of ways = 85C= 83C= 8 x 7 x 63 x 2 x 1= 56
Example 17: In how many ways can 2 boys and 2 girls be selected from a
group of 6 boys and 5 girls?
Ans: Here there are two calculations to be made which are
interdependent.
(i) 2 boys out of 6 boys can be selected in 62Cways and
(ii) 2 girls out of 5 girls can be selected in 52Cways.
The total number of ways of such a selection = 6522 x CC = 6 x 5 2x5 x 42 =150.
Example 18: A book shelf has 10 books of which 6 are of Accounts and
remaining of Management. In how many ways can a person select 3 books
on Accounts and 1 book on Management?
Ans: There are 6 books of Accounts and 10 – 6 = 4 books on
Management.
3 books out of 6 books on Accounts can be selected in 63C= 20 ways.
1 book out of 4 books on Management can be selected in 41C= 4 ways.
By fundamental prin cipal of counting, the total number of ways is 20 x 4 =
80.
Example 19: A Committee of 6 people is to be formed from a Staff of 4
Managers, 6 Officers and 2 peons. Find the number of distinct committees
in which there are: ( i) 2 persons from each category; (ii) no peons; ( iii)
exactly 2 Managers and ( iv) atleast 2 Managers.
Ans:
(i) 2 persons from each category can be selected in
462222 x x CCC= 90 ways.
(ii) No peons to be selected means the selection of 6 people is to be done
from remaining 4 + 6 = 10 people.
This can be done in 106C= 104C= 10 x 9 x 8 x 74 x 3 x 2 x 1= 210 ways.
(iii) Exactly 2 Managers out of 4 can be selected in 42C= 6 ways. munotes.in
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Permuta tion and
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51 The remaining 4 persons in the comm ittee are to be selected from the
remaining 6 + 2 = 8 persons, which can be done in 84C= 8 x 7 x 6 x 54 x 3 x 2 x 1= 70
ways.
The total number of ways = 6 x 70 = 420
(iv) Atleast two managers can be selected in the following ways:
Managers Others
Selection I 2 out of 4 4 out of 8
Selection II 3 out of 4 3 out of 8
Selection III 4 out of 4 2 out of 8
Selection I can be done in 4824 x CC= 420 ways
Selection II can be done in 4833 x CC= 192 ways
Selection II can be done in 4842 x CC= 28 ways
Thus, the total number of ways = 420 + 192 + 28 = 640.
4.7 LET US SUM UP
In this chapter we have learn:
Basic prerequisites and terminology like factorial notation, fund amental
theorem addition and multiplication.
Definition and formula of permutation and some small problems of real life.
Definition and formula of combination and some small problems of real life.
4.8 UNIT END EXERCISE
1. Find the number of ways in which 4 bo ys can be seated for a group
photograph.
2. In how many ways can 3 boys and 2 girls be seated for a photograph.
3. In how many ways can 7 books be arranged on a book shelf?
4. Find the number of ways of making 3 people sit on 3 chairs?
5. Find the number of way s of making 5 people sit on 3 chairs?
6. In how many ways can 3 boys and 2 girls be seated for a photograph in
two lines with the first line of boys and second line of girls?
7. 20 students of a class are seated in three lines for a group photograph.
The firs t line has 8 chairs; second line has 3 chai rs and third 5 chairs.
In how many ways can this be done?
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52 8. In how many ways can 6 people be selected for 3 posts in a company?
9. In how many ways can 10 students be selected for Student’s Council
which consists of only 4 members?
10. Find the number of ways in which a 5 digit number be formed from
the numbers 1, 2, 3, ……, 9 if ( i) no digit is repeated, ( ii) repetition is
allowed.
11. Find the number of ways of forming a 4 digit even number from the
digits 1, 2, 3, ……, 9 if no digit is to be repeated?
12. In how many ways can 3 -digit odd number be formed from the digits
1, 2, 3, ……, 9 if ( i) no digit is repeated, ( ii) repetition is allowed.
13. Find the number of ways of arranging the letters of the word ( i)
MASK,
(ii)MO THER, ( iii) RATION, ( iv) YES, ( v) BHARTI, ( vi)
GREATFUL.
14. Find the value of n from the following:
i. 536nnPP
ii. 6456nnPP
iii. 7512nnPP
15. Show that 11nrP= (n + 1)nrP.
16. Find the number of ways in which 2 books on Mathematics, 3 books
on Law and 2 books on Economics can be arranged on a shelf so that
books of the same subject are together. Also find the number of
arrangements if books are to be arranged at random.
17. Find the number of ways in which 4 books on Physics, 3 books on
Chemistry and 2 books on Biology can be arranged on a shelf so that
(i) books of the same subject are together; ( ii) No two books on
Biology are together; ( iii)No two books on Chemistry are together a nd
(iv) No two books on Physics are together.
18. In how many ways can 4 boys and two girls be seated if ( i) no two
girls sit together, ( ii) no two boys sit together and ( iii) both the girls sit
together?
19. In how many ways can 3 Africans and 3 Americans be se ated so that
(i) atleast two Americans always sit together and ( ii) exactly two
Africans sit together?
20. In how many distinct ways can the letters of the word “CHEMISTRY”
be arranged such that ( i) there is no restriction; ( ii) the word begins
with a vowel; (ii) the word begins and ends with a vowel; ( iii) the
vowels are together and ( iv)the letters T, R and Y are never together. munotes.in
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53 21. In how many distinct ways can the letters of the word “ANDHERI” be
arranged such that ( i) the word begins with A; (ii) the word beg ins with
A and ends with R; (iii) the word begins with vowel and ends with
vowel and ( iv) the letters N, R and I are never together?
22. A College Librarian Mrs. Parita orders 12 books of which 4 are of
Maths, 3 are of English Literature, 3 books on Sociology and 4 books
on Philosophy. In how many different ways can the books be
arranged?
23. In how many distinct ways can the letters of the word “DISMISS” ?
24. In how many distinct ways can the letters of the word “STATISTICS”
such that ( i) no two I’s are together; (ii) no two T’s are together ?
25. Find the number of permutations of the letters of the word “VIRAR”
such that the vowels occupy the even places in the arrangement.
26. In how many ways can 6 books be selected out of 10 books?
27. In how many ways can 4 boys be se lected out of 7 boys?
28. A question paper contains two sections. Section I consists of 4
questions of which two are to be attempted and Section II consists of 5
questions of which 3 are to be attempted. Find the number of ways of
attempting the questions in the paper.
29. There are 6 bulbs of which 3 are defective are to be put in two sockets
in a room. Find the number of ways in which the room is lighted.
30. There are 8 books on History and 4 books on Geography. In how many
ways can 4 books on History and 3 books on Geography be selected?
31. A box contains 4 white and 5 black balls. How many selections of 2
balls can be made so that ( i) one ball of each color is selected; ( ii) both
balls are white; ( iii) no white ball is selected.
32. A box contains 6 blue, 4 green and 2 white balls. How many selections
of three balls can be made so that ( i) one ball of each color is selected;
(ii) atleast one white ball is selected and ( iii) no white ball is selected.
33. 10 candidates appear for an interview. The selection committee has
time to interview only 6 candidates. Find the number of ways of
selecting the candidates.
34. 10 candidates appear for an interview. 2 candidates are disqualified as
they did not bring necessary documents and the selection panel has
time to interview only 5 p eople. In how many ways can this be done? munotes.in
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54 35. A committee consisting of 4 men and 3 women is to be formed from 7
men and 6 women. Find the number of selections in which ( i) a
particular man is selected; ( ii) a particular women is not selected; ( iii)
a particul ar man is selected and a particular is rejected.
36. A College has 8 professors and 5 lecturers. In how many ways can a
committee of 5 teachers be formed such that it consists of ( i) two
professors, ( ii)two lecturers, ( iii) atleast two professors, ( iv) atleas t two
lecturers, ( v)atleast two professors and lecturers and ( vi) no lecturer.
37. A Cultural Committee of 8 persons from 6 men and 5 women is to be
constituted in a Society. Find the number of distinct committees if it
should consist of ( i) atleast 3 women, (ii) exactly 2 women and ( iii)
atmost 4 men.
38. A question paper has three sections with each consisting of 4
questions. A student has to attempt a total of 5 questions with atleast
one from each section. Find the number of ways in which this can be
done.
39. A case is under discussion in front of 5 judges. Find the number of
ways in which the judgment is given with a majority.
40. The Mumbai Royal Cricket Club has 22 players of which 3 are wicket
keepers, 6 are fast bowlers, 2 are spinners and 3 all rounder’s. I n how
many ways can a team be formed if it should include ( i) one wicket
keeper, 2 fast bowlers and one all rounder; ( ii) 2 fast bowlers, 1
spinner and 1 wicket keeper; ( iii) atleast one spinner, ( iv) atleast one
spinner and one all rounder.
Multiple Choic e Questions:
1) If one task can be done in m ways and other task can be done in n ways
then both tasks can be done together in how many ways?
a) 2m + n b) 2mn c) m + n d) m
n
2) How many words can be formed using all the letters of the word
‘GREAT’?
a) 5 b) 24 c) 120 d) 720
3) How many 3 digit numbers can be formed using the digits 6, 7, 8, 9 if
no digit can be repeated?
a) 4 b) 24 c) 48 d) 64
4) 6C4 equals
a) 15 b) 30 c) 120 d)720
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55 5) Number of 3 letter words that can be formed using the letters of the
word ‘DRINK’, exactly once are
a) 60 b) 120 c) 240 d) 480
4.9 LIST OF REFERENCES
Business Mathematics by Qazi Zameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by Hazarika Padmalochan
Business Mathematics by S.R. Arora, Taxmann
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56 5
MATRICES
Unit Structure
5.0 Objective
5.1 Introduction
5.2 Definition of a Matrix
5.3 Order of a Matrix
5.4 Types of Matrices
5.4.1 Equality of two matrices
5.5 Algebra of Matrices
5.6 Let us sum up
5.7 Unit end exercise
5.8 List of References
5.0 OBJE CTIVE
After going through this chapter you will able to know:
Define matrix and its order.
The different types of matrices.
The condition of equality of two matrices.
The operations on matrices.
That to convert real life example in to matrices.
5.1 INTROD UCTION
The theory of matrices was introduced by the English Mathematician
Arthur Gayley in 1858. Since then, a matrix has become a powerful tool to
solve the system of linear equations. In most economic problems we see
the economic variables as line ar functions, although non -linear relations
do exist in practice. ‘Matrix algebra’ enables us to:
(i) write a system of equations in compact form
(ii) test the existence of a unique solution to a system
(iii) solve for equilibrium prices and quantities etc.
Computer analysis of problems involving system of equations applies
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57 computer technology in this field it has become must to study and
understand the basics of matrices and its operati ons.
5.2 DEFINITION OF A MATRIX
A matrix is a rectangular arrangement of numbers (or variables or
parameters) in horizontal rows and vertical columns enclosed in square
bracket [ ] or parenthesis (). As a convention, we represent matrices with
capital l etters A, B, …etc and numbers in it called as “elements” by small
case letters.
Example 1:12 3 401A
The matrix A has 2 rows and 3 columns and hence 2 × 3 = 6 elements viz.
1, 2, 3, -4, 0 and 1.
Example 2: 1 12B
The matrix B has 3 rows and 1 column and hence 3 × 1 = 3 elements.
Example 3: Cx y
Similarly C has 1 row and 2 columns and 1 × 2 = 2 elements.
In general, a matrix with m rows and n columns has m x n number of
elements.
5.3 ORDER OF A MA TRIX
Order of a matrix gives the number of rows and columns in a given
matrix. A matrix is said to be of order m × n (read as m by n) if it has m
rows and n columns.
Referring to the above examples , the order of matrix A is 2 x 3, that of
matrix B is 3 × 1 and that of matrix C is 1 × 2.
Symbolic Representation
A matrix A of order m × n is represented generally as 1 1 ijimAain
Here aij represents element in the ith row and jth column.
Let us understand with an example:
Example 4: Let 74 6 52 1A R2 C1C2C3
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58 The order of matrix Ais 3 × 2 and there are 6 elements.
7 is in the 1st row 1st column, so a11 = 7
–4 is in the 1st row 2nd column, so a12 = –4
6 is in the 1st row 3rd column, so a13 = 6
5 is in the 2nd row 1st column, so a21 = 5
2 is in the 2nd row 2nd column, so a22 = 2
–1 is in the 2nd row 3rd column, so a231 = –1
Thus 12 13 11
22 23 2174 6 = 52 1aa aAaa a
A matrix B of order 4 × 3 would be written symbolically as follows:
13 11 12
23 21 22
4 3
33 31 32
43 41 421 4 = = , 1 3ijb bb
b bb iBbb bb j
b bb
Note that in the notation bij ‘i’ corresponds to the row number and ‘ j’
corresponds to the column number.
5.4 TYPES OF MATRICES
1. Null (Zero) Matrix
A matrix with all elements zero is called as null matrix.
e.g. 1 1 1 32 200 0 , 0 0 0 , 00
2. Row Matrix
A matrix which has only one row is called as a row matrix. Order of a row
matrix is 1 × n.
e.g.1 1 1 3 1 4 1 , 1 1 2 , 725 3
3. Column Matrix
A matrix which has only one column is called as a column matrix. Order
of a column matrix is m × 1.
e.g. 1 1
2 14 11
34 10 , , 59
16
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59 4. Square Matrix
A matrix which has equal number of rows and columns is called a square
matrix. Order of a square matrix is n × n.
e.g. 1 1
2 23 3143
4 , , 3 1 4
431ab
cd
5. Triangular Matrix
(i) Upper triangular matrix: A square matrix whose elements below the
diagonal are zero is ca lled as upper triangular matrix.
e.g.
2 23 312 , 00300ab c
de
f
In above examples; elements {1, 3,} and { a, b, f} are the diagonal
elements.
(ii) Lower triangular matrix: A square matrix whose elements above the
diagonal are zero is called as lower tria ngular matrix.
e.g.
2 23 310030 , 2 6 01275 8
6. Diagonal Matrix
A square matrix whose all elements other than the diagonal are zero is
called as a diagonal matrix.
e.g.13 0 020 , 0 11 0 040 5 17
7. Scalar Matrix
A diagonal matrix whose all diagonal el ements are same is called as scalar
matrix.
e.g.11 0 020 , 0 11 0 020 0 11
8. Identity Matrix
A scalar matrix whose all entries are 1 is called identity matrix and is
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60 e.g.10010 , 0 1 0 01001II
9. Transpose a Matrix
The matrix obtained by interchanging row and column in a given matrix is
called its transpose.
Let A = (aij)m × n be a matrix of order m × n , then its transpose denoted
by AT is given by AT = (aji) n × m
Order of transpose is n × m , which is reverse of m × n as the rows and
columns are interchanged.
e.g. 1 3 3 11 1 2 3 then 2 3TAA
2 3 3 27 1178 1 then 8 13 11 13 2-1 -2TBB
2 236 32 then 24 64TCC
11 12 13 11 21 31T21 22 23 12 22 3231 32 33 13 23 33 then ddd dddDd d d D d d dddd ddd
10. Symmetric Matrix
A square matrix which is same as its transpose is called as symmetric
matrix. Any matrix A is said to be symmetric if A = AT
e.g.41721 , B 1 6 313738A
We can easily check A = AT and B = BT.
11. Skew -symmetric Matrix
A square matrix A is said to be skew -symmetric if A = – AT.
e.g. 02 0 2 then 20 2 0TAA = – AT
also034 0 3 4 3 0 5 then 3 0 5 = 45 0 4 5 0TBB B
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61 5.4.1 Equality of two matrices:
Two matrices are said to be equal if and only if (i) their order is same and
(ii) corresponding elements are same. For equality of matrices both
conditions should be satisfied.
If ijmnAa and ijmnBbandaij = b ij for all i and j, then A = B .
Example 5: Find a, b, x, y if A = B where 57 and 23abABxy
Ans: Since A = B, using the equality of matrices conditions we have,
a = 5, b = 7, x = 2 and y = 3.
Example 6:1123 23
Though the elements may appear to be same, the first condition that the
order of the matrices should be same is not satisfied. Hence the matrices
are not equal.
5.5 ALGEBRA OF MATRICES
1) Addition to Matrices
Two matrices can be added if their order is same and the addition is done
element wise.
If ijmnAa , ijmnBb are matrices of same order m x n then the
sum = ij ijAB a b is also a matrix of order m x n.
Example 7:1 3 1 3 1 1 2 , 1 1 1PQ 1 3 (1 1) ( 1 1) (2 1) 0 0 1PQ
Example 8: 3 1 3 193
= 5 , = 1
24XY
3 1 3 19 3 6
= 5 1 = 6
2 4 6XY
Example 9:2 3 2 311 4 3 2 6 9 = , = 69 2 3 3 7AB munotes.in
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62 2 311 2 4 6 3 9 9 2 12 = = 6 3 9 3 2 7 9 6 9AB
Properties of Addition of Matrices
For any matrices A, B, C of order m x n,
(a) A + B = B + A , matrix addit ion is commutative.
(b) (A + B ) + C = A + (B + C ), matrix addition is associative.
(c) For any matrix, A + O = O + A = A , where O is the null matrix of
same order as that of A.
2) Subtraction of Matrices
Two matrices can be subtracted if their order is sam e and the subtraction is
done element wise.
If ijmnAa, ijmnBbare matrices of order m x n then = ij ijmnAB a b
Example 10:46 0 2 = = 31 11AB 4 0 6 2 4 4 = 3 ( 1 )1 1 40AB
3) Scalar Multiplication of a Ma trix
A scalar is any constant (number). Let ijmnAa and α be any scalar
Then = ijmnAa
When a matrix is multiplied by a scalar, every element of the matrix gets
multiplied by that scalar.
Example 11: 7 4 then 2 14 8CC
Example 12: 5 3 2 then 4 20 12 8AA
Example 13: 46 231 then 282 41BB
Properties of scalar multiplication
Let A, B matrices of same order and
be scalars:
(a)
(A + B ) =
A +
B
(b) (
+
) A =
A +
A munotes.in
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63 (c)
(
A) =
(
A) = (
Before we proceed to next section, let us understand the above properties
and with simple examples.
Example 14: Write down the order of each of the following matrix. State
which pair of matrices can be added and write the resultant matrix.
26 2 = 6 8 , = , = 4 , = 2 4 , = , = 2 , 55 4
1445 4 6 1 2 11 = , = , = , = 4 7 1 , 23 4 4 3 8717AB C D E F
GH J K
Ans: The o rder of a matrix is m × n, where m = no. of rows and n = no. of
columns. Thus, the order o f given matrices are:
Matrix Order Matrix Order
A 1 × 2 E 2 × 2
B 2 × 1 F 1 × 1
C 1 × 1 G 2 × 3
D 1 × 2 H 2 × 2
J 2 × 1 K 3 × 3
Now, we know that two matrices can be added only if their order is same.
Thus, from the above table, we can conclude that the following matrices
can be added:
AandD, BandI, CandF, EandH and their sums are:
A + D = 6 8 + 2 4 = 6 2 8 4 = 4 12
B + J = 21 12 1 11 3 + = = 58 5 8 3
C + F = 2 + 2 = 4 2 = 6
E + H = 62 12 6 12 2 54 + = = 54 4 3 5 44 3 97
Example 15: Verify the commutati ve and associative laws of addition for 31 3 2 43 = , = , = 42 41 21ABC
Ans: To verify: munotes.in
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64 (i) Commutative law: ABBABCCBAC CA
(ii) Associative law: ( ) ( ) AB C A B C
(i) We will verify here A + B = B + A. The other two being similar is left
for readers as exercise!
A+ B = 3 1 3 2 3 3 1 2 0 1 + = = 42 41 4 42 1 8 3 … (1)
B + A = 3 2 3 1 3 3 2 1 0 1 + = = 41 42 4 41 2 8 3 … (2)
From (1) and (2), A + B = B + A. Hence commutative law holds true.
(ii) B + C = 3 2 4 3 3 4 2 3 1 1 + = = 41 21 4 21 1 62 A + (B + C) = 3 1 1 1 3 1 1 1 4 2 + = = 42 62 4 62 2 1 04 … (3)
Now, A + B = 3 1 3 2 3 3 1 2 0 1 + = = 42 41 4 42 1 8 3
(A + B) + C = 01 4 3 0 41 3 4 2 + = = 8 3 2 1 8 2 3 1 10 4 … (4)
From (3) and (4), A + (B + C) = (A+ B) + C
Thus, Associative law holds true.
Example 16:Write down elements of a matrix A of order 2 × 3, if 2 ( )ijai j .
Ans: Given matrix ()ijAa of order 2 × 3 i.e. containing 6 elements. 12 131122 2321 Symbolically, = aaaAaaa … (*) Also
given is the formula to calculate eachijai.e. 2 ( )ijai j
22
11
22
12
22
13
22
21
22
22
22
23 (1 1) 2 4 (1 2) 3 9 (1 3) 4 19 (2 1) 3 9 (2 2) 4 16 (2 3) 5 25a
a
a
a
a
a
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65 Substituting these values in (*), we have A = 49 1 6 91 62 5
Example 17: Find a matrix B such that A + B = O, where A = 46 8 32 7
Ans: Let us first recollect that two matrices can be added if their order is
same and the resultant matrix is also of the same order.
Given matrix A has order 2 × 3.
B and 0 also must be order 2 × 3. As we know 0 is the Null Matrix, we
have
A + B = 0
B = 0 – A = 000 4 6 8 000 3 2 7
B = 46 8 32 7
Example 18: Find a matrix such that 2A + 3B = O where A = 3 129182.
Ans: As we have done in previous example, we know that B and O are
matrices of order 2 × 2.
Given 2A + 2B = O
3B = O – 2A = 3 1200 2 918002
3B = 0 0 6 24 6 24 00 9 3 6 9 3 6
B = 6 2462 4133 39 36 9 3633
B = 2831 2
Example 19: Find all 2 × 2 matrices A such that A + AT = 0 (in other
words, A is a skew -symmetric matrix)
Ans: Since A is a 2 × 2 matrix, writing A symboli cally, munotes.in
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66 A = 11 1221 22aaaa AT = 11 2112 22aaaa
Given A + AT = 0
11 12 11 21
21 22 12 2200 00aa aa
aa aa
11 11 12 21
21 12 22 22 00 00aa aa
aa aa
11 12 21
21 12 222 00 200aa a
aa a
11 11 2 0 0aa 12 21 12 21 + 0 aa a a
and22 222 0 0aa 21120 0aAa
Thus, only 2 × 2 skew -symmetric matrix is of the form: 00xx
WORD PROBLEMS
Example 20: Three friends Amit, Sumit and Vinit go to a retail shop and
purchase following items :Amit purchase s one mango, 2 kg potatoes and 1
kg rice. Sumit purchases half a dozen lemons, 1 kg potato and 1 kg sugar.
Vinit purchases one dozen lemon, 3 kg rice and 1 kg sugar. Construct a
proper matrix giving various purchases of the three friends.
Ans: There are t hree friends which can be taken as three rows. And there
are 5 items which can be taken as five columns. Thus we gave the matrix
of purchase as:
Mangoes Potatoes Rice Lemon Sugar
Amit 12 2 1 0 0
Sumit 0 1 0 6 1
Vinit 0 0 3 12 1
3 512 2 1 0 0
0 1 0 6 1
00 3 1 2 1A
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67 Example 21: A multinational bank has its offices in Mumbai and Delhi. In
both the cities it has 3 branches. Every branch has 1 manager, 4 officers
and 3 peons. In one of Mumbai branch they have 2 executives, while in all
the Delhi branches there are 5 typists. Use matrix notation and find ( i)
total number of posts of each type in all branches together ( ii) total
number of employees of each type is both offices taken together.
Ans. Consider the following rows matrices: 123 1 4 3 5 0 , 1 4 3 0 0 , 1 4 3 0 0MMM
123 1 4 3 0 5 , 1 4 3 0 5 , 1 4 3 0 5DDD 123 , , MMMrepresents branches in Mumbai, elements represent number of
manager, officers, peons, executives and typists in the order of
appearance. Similarly 123 , , DDD represent the three branches in De lhi
(i) Total number of posts of each type
(a) in Mumbai : 123 , , MMM 1 4 3 2 0 1 4 3 0 0 1 4 3 0 0 3 12 9 2 0
(b) in Delhi : 123 , , DDD 1 4 3 0 5 1 4 3 0 5 1 4 3 0 5 3 1 2 9 0 1 5
(ii) Total number of emp loyees in both offices together is
(123 , , MMM) + (123 , , DDD) 3 12 9 2 0 3 12 9 0 15 6 24 18 2 15
Example 22: An automobile company sells two models of a car AC and
Non AC, made available in three colors silver, black and red. In the first
month of launching the sales for AC version were 120, 180, 205 for silver,
black and red color respectively, while for the Non AC version the
corresponding numbers were 220, 350, 310 respectively. In the second
month th e sales for AC and Non AC Models were: 360, 275, 327, and 530,
320, 382 respectively. Represent the above data in matrix from and find
the total sales for each model and color for both the months together.
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68 Ans. Les us represent the given data in matrix form:
1st Month 2nd Month
AC Non AC AC Non AC
Silver 120 220 360 530
Black 180 350 275 320
Red 205 310 327 382
Let A = 120 220180 350205 310 and B = 360 530275 320327 382 be matrix representing sales in
the two months respe ctively.
The total sales can be obtained by adding A and B.
A + B = 120 220 360 530 480 750180 350 275 320 455 670205 310 327 382 532 692
4) Multiplication of Two matrices:
Two matrices A and B are said to conformable or in other words they can
be multiplied, if number of columns of A is same as nu mber of rows of
B. The product AB makes sense only if the above condition is satisfied.
Let A = [ aij] be a m × n matrix and B = [ bjk] be a n × p matrix. Then the
product AB is defined as:
AB = [ cik], where cjk = a i1b1k + a i2b2k + ….. + a inbnk and order o f AB is m
× p.
Note:
1. Observe that product AB has number of rows that of A and number of
columns that of B. x = mn np mpA B AB
2. The multiplication of matrices is not element wise as in addition or
subtraction of matrices.
As mentioned the above matri x AB = [ cik] is of order m × p.
Here cik = a i1b1k + a i2b2k + ….. + a inbnkis the ikthelement of AB, obtained
by multiplying the row of A with column of B.
Let us understand this with examples of square matrices order 2 × 2 and
3 × 3:
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69 Example 23: Let A = 11 1221 22aaaa , B = 11 1221 22bbbb be matrices of order 2 × 2.
Thus product AB also is a matrix of order 2 × 2.
Let AB = 11 1221 22cccc
The four elements 11 12 21 22 , , and ccc c are calculated as follows:
1. c11 is obtained by multiplying 1st row of A with 1st column of B i.e. 11 12 aa with 1112bb 12 11 11 12 21 ca b a b
2. c12 is obtained by multiplying 1st row of A with 2nd column of B i.e.
1112aa with 1221bb 12 11 12 12 22 ca b a b
3. c21 is obtained by multiplying the 2nd row of A with 1st row of B. 21 21 11 22 21 ca b a b
4. Similarly,22 21 12 22 22 ca b a b
Thus, AB = 11 11 12 21 11 12 12 2221 11 22 21 21 12 22 22
ab ab ab abab ab ab ab
Example 24:Let A = 1432 , B = 5231
11 1 5 ( 4) (-3) 5 12 17C
12
21
22 1 2 ( 4) 1 2 4 2 3 5 2 ( 3 ) 1 5 6 9 3 2 2 1 6 2 8C
C
C
AB = 17 298
It will take a few problems of practice to get accustomed to multiplication
of matrices. L es us revise again the method of how did we get the
elements of first row in AB:
It is by multiplying 1st row of A with 1st column of B and 1st row of A by
2nd Column B. While the elements in 2nd row of AB, again, by multiplying
2nd row of A with 1st column of B and 2nd row of A with 2nd column of B. munotes.in
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70 Example 25: For A = 3 273
1 4
25B = 2 363 0 42 1 find AB
Ans: A and B are conformable as no. of columns of A = no. of rows of
B = 2
and AB has order 3 × 3.
Let AB = 11 12 1321 22 2331 32 33ccccccccc
C11 is obtained by adding the products of elements in 1st row of A and 1st
column of B.
c11 = 7 x 6 + 3 ( – 4) = 42 – 12 = 30
c12 = 7 ( –3) + 3 x 2 = –21 + 6 = –15
c13 = 7 x 0 + 3 ( – 1) = –3
c21 = 1 x 6 + 4( –4) = 6 – 16 = –10
c22 = 1(–3) + 4 x 2 = 5
c23 = 1 x 0 + 4( –1) = –4
c31 = 2 x 6 + 5( –4) = –8
c32 = 2(–3) + 5 x 2 = 4
c33 = 2 x 0 + 5( –1) = –5
Thus, AB = 30 15 310 5 484 5
Example 26: A = 3122 , B = 2406 , find AB and BA.
Ans. Firstly A and B are matrices of order 2 × 2. Hence AB and BA both
products are defined.
BA = 2 4 3 1 ( 2) 3 4 ( 2) ( 2) 1 4 2 0 6 2 2 0 3 6 ( 2) 0 1 6 2
BA = 14 612 12 … (1)
And AB = 31 2 4 22 0 6 munotes.in
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71 3 ( 2) 1 0 3 4 1 6 ( 2) ( 2) 2 0 ( 2) 4 2 6
BA = 61 844 … (2)
From (1) and (2) it is very clear that AB ≠ BA. This is very important fact
regarding product of matrices. Matrix multiplication in general is not
commutative. i.e. AB ≠ BA for any matrices A and B.
There are some examples, as we see further, where AB = BA, but as a
rule, it is not true always.
Example 27: For A = 1000 , B = 0001 show that AB = BA.
Ans: AB = 10 00 00 = ...100 01 00
BA = 00 10 00 = ...201 00 00
From …1 and …2 AB = BA
Example 28:Let A = 11 8 , B = 36 , C = 0112 , D = 11 6 78 2
Can you compute AB, BA, AC, OB, DA, CD, AD? If yes ,find it or else
justify your answer.
Ans: Given Matrix Order
A 1 × 2
B 2 × 1
C 2 × 2
D 2 × 3
We know that two are matrices conformable if number of columns of first
matrix is same as number of rows of second matrix. From the above table,
we can say that A and B, B and A, A and C, C and D, A and D are
confo rmable. Matrices D and B, D and A are not conformable.
DB and DA cannot be computed.
(i) AB = 1 211 82 13
6= 1 111 3 ( 8) ( 6) = 33 48
AB = [81] munotes.in
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72 BA = 2 13
62 211 8
BA = 2 23 11 3 ( 8)
6 11 ( 6) ( 8) = 33 2466 48
(ii) AC = 11 82 201
12
=1 211 0 ( 8) ( 1) 11 1 + ( 8) 2
AC = 85
(iii) CD = 2 201
122 311 6 78 2= 2 30 1 1 7 0 ( 1) 1 ( 8) 0 6 1 2 1 1 2 7 ( 1) ( 1) 2 ( 8) ( 1) 6 2 2
CD = 78 2 13 15 2
(iv) AD = 1 211 82 311 6 78 2
= 1 311 1 ( 8) 7 11 ( 1) ( 8) ( 8) 11 6 + ( 8)x 2
AD = 1 345 53 50
Properties of Matrix Multiplication:
(1) Law of Commutativity do es not hold true: For any two conformable
matrices A, B; AB ≠ BA
(2) Law of Associativity: A(BC) = (AB) C
(3) Existence of Multiplicative Identity: AI = IA = A , where I is the
identity matrix.
(4) Law of cancellation does not hold true: If AB = AC , it does not B = C.
(5) If AB = 0, it does not imply A= 0 or B = 0 (Refer to Ex. 5)
(6) Positive integer powers of a matrix:
A2 = A.A.; A3 = A × A × A
In general An = A × A × ….. × A n times.
(7) Law of Distributivity: For conformable matrices A, B, C;
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73 (8) In view of property (1), we have some more results:
(i) (A + B) 2 = A2 + AB + BA + B2 (and not A2 + 2AB + B2)
(ii) (A – B) 2 = A2 – AB – BA + B2
(iii) (A + B) (A – B) = A2 – AB + BA – B2
Example 29:If A = 1121 and B = 1141 then show that
(A + B) 2 = A 2 + B 2
Ans: We know that (A + B) 2 = A2 + AB + BA + B2 … (1)
To prove (A + B) 2 = A2 + B2 … (2)
it is enough to prove AB + BA = 0 … from (1) & (2)
AB = 11 1 1 1 1 ( 1 ) 4 1 1 ( 1 ) ( 1 ) = 2 1 4 1 2 1 ( 1) 4 2 1 ( 1) ( 1)
AB = 3223 … (3)
BA = 11 1 1 1 1 1 2 1 ( 1 ) 1 ( 1 ) = 4121 4 1 ( 1 ) 2 4 ( 1 ) ( 1 ) ( 1 )
BA = 3223 … (4)
Adding (3) & (4), we have
AB + BA = 32 3 2 00 + = = 023 2 3 00
Thus (A + B) 2 = A2 + B2
Example 30:If A = 4213 , compute A2 – 8A + 6I
Ans: A2 = 42 42 1 6 28 6 1 8 1 4 = = . . . 113 13 4 3 2 9 71 1
8A = 8 42 3 21 6 = ... 213 8 2 4
6I = 6 10 60 = ... 301 06
from (1), (2) and (3), we have munotes.in
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74 A2 – 8A + 6I = 18 14 32 16 6 0 71 1 8 2 4 0 6
= 18 32 6 14 16 0 8 2 = 7 8 0 11 24 6 1 7
Word Problems
Example 31:Mr. Narayan buys 3 kg potatoes, 1 kg onion, 5 kg rice, 1 kg
wheat and 2 kg oil from a shop. If the cost per kg of above items is Rs.
10/-, Rs. 12/ -, Rs. 18/ -, Rs. 15/ - and Rs. 50/ - respectively, find using
matrix algebra the total amount spent by him.
Ans. Let A = 3 1 5 1 2 represent the items bought by Mr. Narayan
and B = 10 12 18 15 50T represent the cost per kg of items.
Then the total expenditure = AB = 10123 1 5 1 2 181550
AB = 3 10 1 12 5 18 1 15 2 50
AB = 30 12 90 15 100 = 247
Total amount spent = Rs. 247/ -.
Example 32:A company produces three types of products say A, B, C. Let
the matrix X represent the sales of these products in two cities as follows:
X = 10 40 3020 30 40If manufacturing cost of the products is Rs. 100/ -, Rs.
150/- and Rs. 250/ - respectively and is sold at Rs. 120/ -, Rs. 140/ - and Rs.
300/- respectively, find the total profit using matrix algebra.
Ans: Profit = Selling Price – Cost Price
Let C = 100 150 250 and S = 120 140 300
Represent the cost price and selling price per product of type A, B, C.
Total manufacturing cost :
Consider CXT = 10 20100 150 250 40 3030 40
CXT= 1000 6000 7500 2000 4500 10000
CXT = 14500 16500 munotes.in
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75 Total manufacturing cost = 14500 + 16500 = Rs. 31,000/ - …1
Total Selling price
Consider SXT = 10 20120 140 300 40 3030 40
SXT = 1200 5600 9000 2400 4200 12000
SXT = 15800 18600
Total money after sales = 15800 + 1860 0 = Rs. 34,400/ - …2
From (1) and (2)
Profit = 34400 – 310000 = Rs. 3400/ -
Example 33:Mr. Mittal owns two factories A and B which produces two
products P and Q. Factory A produces 12 units of P and 6 units of Q and
factory B produces 14 units of P and 5 units of Q, represented by the
matrix:
X = 12 1465 . If Y = 45 represents number of days the two factories
operate, what does XY represent ?
Ans. Y = 45 means, factory A operates for 4 days a week and factory B
5 days a week.
XY = 12 14 4 12 4 14 5 48 70 = = 65 5 6 4 5 5 2 4 2 5
XY = 11849 represents number of units of P produced in both the
factories in a week (ie 118) and number of units of Q produced in both the
factories in a week (ie 49).
5.6 LET US SUM UP
In this chapter we have learn:
Definition and order of matrix.
Types of matrices.
Equality of two matrices.
Operations on matrices.
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76 5.7 UNIT END EXERCISE
1) Write down the order of the following matrices:
(a) 24368 , (b) 479 586 , (c) 222222xx y yy yz zzx z x , (d) abc , (e) [100]
2) 11 4 16 5 2 4 1 9 16 42 29 , 2 1 0 1 6 , 9 2 1 116 9 33 4 16 15 16 22 3AB C
Verify the following matrices:
(a) A + B = B + A, (b) A + (B + C) = (A + B) + C, (c) A + C = C + A,
(d) A – (B + C) = A – B – C, (e) A – (B – C) = A – B + C
3) For 67 3 3 , 43 7 6AB
Verify:
(i) (A + B) T = AT + BT
(ii) (AT) T = A
4) If A = [a ij]3 × 3 is such that a ij = i – j . Find all elements of A and write
down the matrix A.
5) If 34 46 , 71 1 1 2 1 0PQ . Find 3P – 2Q.
6) (i) Find the matrix B if 4A + 5B = 3I, where A = 2112 and I is
the identity matrix.
(ii) If41 3 4 8 1 1 , , 23 0 1 6 5AB C . Find the matrix X such
that 2A – 3B + 4C = 5X.
7) If 4 2 8 6xy xz
zy xy , find x, y, z.
8) If A = 11 4576xzy is symmetric matrix, find x, y, z. munotes.in
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77 9) If A = 050160ab
c is a skew -symmetric matrix, find a, b, c.
10) Find the values of x, y, z if 17 31 2 2 965 4 3 2 1 311 3 6 5 4 17 4 8xx
yz
zy
11) If, 6 8 6 7 7 12xy
xy find x, y.
12) Find a, b, c and d, if
(i) 2 3 3 1 1 3 1 4 3 4 1 1 6 1 0ab c a
ab cd
(ii) 5 41 3 4 3 7 4 1 7ab c a a d
ac
13) Solve the following to find matrices P and Q.
(i) 44 4 43 , 3 3 1 1 21PQ P Q
(ii) 415 3 3 0 2 , 2 14 4 3 3 2PQ P Q
(iii) 11 0 12 , 3 01 1 0PQ P Q
14)
(i) If A = 1023 , find a matrix B such that A + B is a diagonal matrix.
(ii) If A = 102431214 , find a matrix B such that A – B is scalar matrix.
(iii)If A = 123 100 , find a matrix B, such that A + B is a Null matrix.
(iv) If A = 12 376 489 5 , find a matrix B such that A + B is an upper
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78 (v) If A = 16 728 934 5 , find a matrix B such that A + B is a lower
triangular matrix.
15) From a grocery store Ram purchased 3 kilo potatoes, 1 kilo onions,
and 2 kilo rice. Shyam also purchased 2 kilo potatoes, 1 kilo rice and half
a kilo wheat. Represent the above data in matrix form.
16) The number of students in Arts, Science and Commerce faculty
studying in F .Y., S.Y. and T.Y. of a certain college of as follows: In Arts
there are 120, 60 and 25, in Science there are 440, 280 and 120 and in
Commerce there are 720, 500 and 300 students in F.Y., S.Y. and T.Y.
respectively. Represent the above data in matrix form.
17) In an entrance examination, 20 students from college A, students from
college B and 35 from college C appeared. Only 12 students from A, B
and 10 students from C passed the examination. Out of them 5 students
from A, 7 students from B and 8 from C se cured 100% marks. Represents
the above data in matrix from.
18) Let A =12 14 1513 15 1714 16 15, where the three rows represent number of three
items and the columns represent three shops respectively. If B =0112 1 2130
, represent the s ales per day, find the stock left after the sales done in five
days using matrix notation.
19) In a city there are 12 colleges and 52 schools. Each institution has 6
peons, 4 clerks, and 1 cashier. Each college in addition to this as 1 head
clerk and 1 ac countant. Using matrix method, find the total number of
posts in school and colleges taken together.
20) The number of students enrolled for a personality contest from three
colleges A, B and C were 15, 24 and 8 respectively. Of these those who
enter the semi -final were 8, 16 and 4 respectively. Only 4, 9 and 1 students
from the three colleges respectively qualified for the finals. Represent the
above data in matrix form.
21) The price of one kilo of rice, wheat and oil in a retail shop is Rs. 16/ -,
Rs. 12/- and Rs. 65/ - respectively. The same items costs Rs. 15/ -, Rs. 10/ -
and Rs. 50/ - in a wholesale shop respectively. Ramesh and Suresh
purchased 5 kilos of each items from the retail and whole sale shop
respectively. Using matrix method, find the total c ost incurred by each of
them.
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79 22) A firm manufacturers three items; A, B and C. The sales in first
quarter are 1.18, 2.25 and 3.02 (in million tons) respectively, while that in
the second quarter are 1.06, 2.40, 3.01. Find the half yearly sales for the
three items.
23) Which of the following matrices are conformable?
(i) A = 100 9075 32 , (ii) B = 1001 1101 1201 11 101 1001 , (iii) C = 345 ,
(iv) D = 6 2 1141 41 13 6 , (v) E = 14 10 9
24) If A = 4- 63 12 9 , B = 416 239 , compute AB, BA. Is AB = BA
justify your answer.
25) (i) Let P = 4012 , Q = 0231 , R = 1142 . Compute PQ,
PR, QR, P(QR), (PQ)R. What can you say about P(QR) and (PQ)R ?
(ii) For above matrices, P, Q, R,; compute (P + Q) R, (P + Q) R. Hence
verify P(Q + R) = PQ + PR, (P + Q) R = PR + QR.
26) If A = 113 333 , B = 315 224 , C = 403 552 . Verify that AB = AC
but B ≠ C.
27) If f(x) = x2 – 3x + 6, find f(A) where A = 2312 .
28) Verify that AB = 0, where A =12 011 0140 , B = 000000149
29) If A 12 3 3 3 3 = 456 4 5 6 find A.
(Hint : Assume A to be a matrix of order 2 × 2 and solve.)
30) If A = 4112 , find A + 5A – 10I.
31) Verify that A = 2312 satisfies A2 – 4A + I = 0 munotes.in
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80 32) Find matrix x such that 3123X = 01 15 11 11 18
33) If A = 3242 , find the value of x such that A2 – xA + 2I = 0
34) If A = 1004 , B = 3002 , show that AB = BA.
Prove that, a diagonal matrices commute under multi plication.
35) For matrices P =321 , Q = 12 3 , verify that (PQ) T = QTPT.
36) If X= 4211 , show that ( X – 2I) (X – 3I) = 0
37) If A = 21 123 244 3 show that A (A –
38) A sof t drink company manufactures two types of soft drinks A and B.
The sales of these products in two cities are represented by the matrix X as
follows:
A B
X = 1000 1200 1500 2500
If Y = 1210 represents the cost of each produc t A and B respectively, what
does XY represent.
39) A manufacturer produces three products A, B and C and sells it in two
cities the total sales are shown as follows:
AB C13500 11000 4500 City 18200 15000 7800 City 2
(i) If the selling price per unit is Rs. 4/ -, Rs. 6/ - and Rs. 3/- for the product
A, B and C respectively, find the total revenue using matrix algebra.
(ii) If the manufacturing cost per unit is Rs. 2/ -, Rs. 3/ - and Rs. 0.50/ - for
the product A, B and C respectively, find the overall profit using matrix
algebra. City 1
City 2 munotes.in
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81 40) The sales of three items of a company from Monday to Saturday is
represented by the matrix X as follows:
X = Mon Tue Wed Thu Fri Sat2 5 3 1 4 6 Item 1 4 3 1 2 5 6 Item 26 5 2 3 1 4 Item 3
Let P = 213 be the matrix representing the price of each item respectively.
(i) What does 111111XTrepresent ?
(ii) What does XT111 represent?
(iii) Find the total amount obtained at the end of week.
41) The sales of three brands of Tea A, B, C during a week from Monday
to Saturday are shown below by matrix X.
X = Mon Tue Wed Thu Fri Sat10 15 12 11 10 20 A 20 20 30 30 25 30 B5 10 8 7 5 10 C
Let C = 213 be the cost per packet of Tea brand respectively.
(i) Find total sales of the three brands during a week.
(ii) Find the total earnings in a week.
(iii)Find total sales in a week of all bra nds together per day.
(iv) Find total sales during a week of all brands taken together.
42) ‘Nirman Constructions’ has taken a order to construct 10 houses of
type A, 20 houses of type B and 5 houses of type C. The following matrix
represents the raw mater ial required in each type of house.
X = 105 50 70 20 100 A150 70 50 15 120 B 75 35 20 10 50 CCement Steel Wood Glass Paint
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82 The cost per unit raw material is given by Y = [ 10 18 6 15 9]
(i) Find how much raw material the company should order.
(ii) Find cost of all materials for a house of each typ e.
(iii)Find total cost of raw materials for all houses taken together.
5.8 LIST OF REFERENCES
Business Mathematics by QaziZameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
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83 6
DETERMINANT
Unit Structure
6.0 Objective
6.1 Introduction
6.2 Determinant of matrix of order 2x2
6.3 Determinant of matrix of order 3x3
6.4 Properties of Determinants
6.5 Cramer’s rule
6.6 Let us sum up
6.7 Unit end exercise
6.8 List of Referenc es
6.0 OB JECTIVE
After going through this chapter you will able to know:
Determinant of square matrices of order upto
Properties of determinants.
Using determinant able to solve system of equations.
6.1 INTRODUCTION
Determinant is a value asso ciated with a square matrix. Let A be a square
matrix then determinant of A is denoted as | A|.
As we know now that determinant is defined for square matrices only, let
us learn how to compute determinants of square matrices of order 2x2 and
3x3.
6.2 DETERM INANT O F MATRIX OF ORDER 2X2
Let []ijAa be a matrix of order 2x2. Explicitly we can write 11 1221 22aaAaa.
Then, 11a 12a
|A| = 21a 22a = 11a22a– 12a21a (*)
Note : The minus sign between the two terms is the part of the formula. munotes.in
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84 Example 1.If 2437A, then 2 4||3 7A= 2 x 7 – 4 x 3 = 14 – 12 = 2
Example 2. Let 3162A
Then, 31||6 2A= 3 x ( –2) – (–1) x 6 = –6 + 6 = 0
Example 3.If 11 23 x= 50, then find the va lue of x.
Ans: Evaluating the determinant on the LHS, we have,
11x – (–2) x 3 = 50
i.e.11 x + 6 = 50
11x = 50 – 6 = 44
x = 4
6.3 DET ERMINANT OF MATRIX OF ORDER 3X3
Let A = [aij] be a matrix of order 3x3. Explicitly we can write A
as:11 12 1321 22 2331 32 33aaaAa a aaaa11 12 1322 23 21 23 21 2221 22 23 11 12 1332 33 31 33 31 3231 32 33aaaaa aa aaAa a a a a aaa aa aaaaa
Note : There are three terms on the RHS of the formula
1. The first element a11 is multiplied with the minor determinant of order
2x2 obtained by del eting the row and column containing a11.
2. The second element a12 taken with a minus sign is multiplied with its
minor determinant of order 2x2 obtained by deleting the row and column
containing a12
3. The third element a13 is multiplied with its minor determ inant as
described above for the first two terms.
Initially readers are advised to use the above formula to evaluate
determinants of order 3x3. After gaining confidence one can write the
same directly as follows: 11 22 33 23 32 12 21 33 23 31 13 21 32 22 31() () ()Aa a a a a aa a a a aa a a a munotes.in
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85 Example 4. Evalu ate the determinant 123245314
Ans: 12345 25 24245 1 2 314 34 31314
= 1 (4 x 4 – 5 x 1) – 2(2 x 4 – 5 x 3) + 3(2 x 1 – 4 x 3)
= 11 – 2(– 7) + 3( – 10) = 11 + 14 – 30
= – 5
6.4 PROPERTIES OF DETERMINANTS
Determinants satisfy certain important properties which makes it more
useful in solving complex problems. Here we will be discussing some
elementary properties of determinant.
1. The value of determinant remains unchanged if its rows and
columns are interchanged. In other words the dete rminant of a matrix is
same as that of its transpose, || | |TAA. For example,
4585312
And 4185352
2. If any two rows (or columns) are interchanged then the value of
determinant changes only by sign. For example,
4185352. Now if we interchange the first and second row,
i.e. 12RR gives 5258 341
Or if we interchange the columns of the original determinant i.e12CC
5458 321
3. If any two rows (or columns) are identical or if one row (column) is
a multiple of another row (column) then the value of determinant is zero.
For example, 4244021 … (as R1 = 2R2.)
Also, 4422 = 8 – 8 = 0 … (as C1 = C2)
4. If any row (or column) is multiplied by a scalar, say then the value
of determinant is also times the original value i.e. ||| |AA. munotes.in
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86 For example, 3124= 12 + 2 = 14 and if we multiply first row by 2, then
the determinant i.e 2 R1gives 6224= 24 + 4 = 28 = 2(14)
Similarly, 2 3 03 0 4 2(0 16) 3(15 0) 0 32 45 770 4 5
Then 3 C3 gives, 2 3 03 0 12 2(0 48) 3(45 0) 0 96 135 231 3( 77)0 4 15
5. Any operation of the type ( or )ij i jRk R Ck C does not change the
value of the determinant. The meaning of the row operation is that to R1
we add k times R2. For example, 2131= 2 + 3 = 5
Now to R1 if we add 2 R2 i.e.R 1 + 2R 2 then the determinant becomes 2 2(3) 1 2(1) 8 131 3 1 = 8 – 3 = 5 (same !!)
This is the most powerful property of determinants which is extensively
used in solving determinants and inverses of matrices.
Let us solve some problems using the above properties of determinants.
Example 1. Evaluate the determinant without actually expanding it 1 11 1211 12 1441 13 169.
Ans: Consider 1 11 1211 12 1441 13 1691 11 1211 12 1440 1 25 (by the operation R3 – R2 )
1 11 1210 1 230 1 25 (by the operation R2 – R1)
1 11 1210 1 2300 2 (by the operation R3 – R2 )
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Determinant
87 10 011 1 0121 23 2 (by interchanging rows to columns)
= 1(2 – 0 ) – 0 + 0
= 2
Example 2. Without expanding the determinant show that
12 6 4 304 3 12 6
Ans: Now (by the property 2) we know that i f any two rows (or columns)
are interchanged then the determinant changes by a sign,
L.H.S. = 12 6 4 343 1 26 = 43 4312 6 12 6 (by the operation 12RR)
= 0 = R.H.S.
Example 3. Without expanding the determinant show that 12456789xx x
xxx
xxx= 0
Ans: L.H.S. = 12456789xx x
xxx
xxx
= 1145 178 1xx
xx
xx
(by the operation C3 – C2)
= 11411711x
x
x
(by the operation C2 – C1)
= 0 Qif any two rows ( or columns ) are identical then the determinant is zero.
In this problem C3 = C2.
Example 4. Without expanding show that ( )( )( )abcbac ab ac abcbca .
Ans: Consider munotes.in
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88 L.H.S.=abcbacbcaabbaccbacbca() 0ab abba cbc a
(by the operation R1 - R2 ) 11 0()ab bbaccabca 11 0() 0 ()ab ac acbc a
(by the operation R2 – R3) 11 0( )( ) 0 1 1ab acbc a 10 0( )( ) 0 1 1ab acbcb a
(by the operation C1 + C2)
= (a – b)(a – c)[1(a + b + c ) – 0 + 0 ] = (a – b)(a – c)(a + b + c ) = R.H.S.
6.5 CRAMER’S RULE
System of Linear Equations :
In this section we shall discuss the method to solve a system of linear
equations using Cramer’s rule.
Consider a system of two linear equations in two variables : 11 122 2ax by cax by c; the solution of these equations is given by 11 1 1 1122 2 2 22 , where D= , and y x
xyDab c b acDxy D Dab c b acDD
This is called as Cramer’s rule.
Consider the three equations in three variables : 111 1222 233 2 3ax by cz dax by cz dax by cz d; the solution of these equations is given by , and ,y xzD DDxy zDD D 11 1 11 1 1 1 1 111222 222 2 22 22 2333 333 3 33 333 where D , , and xy zab c db c adc ab dabc D dbc D adc D abdabc dbc adc abd
The Cramer’s rule is applicable only when0D. munotes.in
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89 Let us solve some system of linear equations using Cramer’s rule
Example 1.235xyand22xy
Ans: The de terminant of the coefficients is 2 31 2D = 4 – 3 = 1
Also, 5310 6 422xD and 2545 112yD
By Cramer’s rule we have,
,yxDDxyDD
441x and111y
Hence th e solution of the given system of equations is
x = 4 and y = – 1
Example 2.23 9xy z , 26xy z and 32 8xy z
Ans: We have 2311 2 1 2(4 1) 3(2 3) 1(1 6) 6 3 5 4312D
9316 2 1 9(4 1) 3(12 8) 1(6 16) 27 12 10 5812xD
2911 6 1 2(12 8) 9(2 3) 1(8 18) 8 9 10 7382yD
2391 2 6 2(16 6) 3(8 18) 9(1 6) 20 30 45 5318zD 4, 5, 7, 5xy z DD D D
By Cramer’s rule we have , and ,y xzD DDxy zDD D
54x, 74y and54z.
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90 Example 3. In a factory two products are produced, A and B using two
machines
M1 and M 2.The time taken by product A to be processed through M 1 and
M2 is 2 and 1 hour respectively. The time taken by product B to be
processed through M 1 and M 2 is 1 and 2 hours respectively. If the time
available for both the mach ines is 30 hours, how many units of the two
products can be produced?
Ans: Let x units of product A and y units of product B be produced. Then
from the given information we have the following system of equations:
2 30xy …. (1)
2 30xy …. (2)
We solve the above linear equations using Cramer’s rule.
2141312D 3D.
30 160 30 3030 2xD 30xD
23 060 30 3013 0yD 30yD
Now w e know that ,yxDDxyDD .
Thus, 30 30=10 and 1033xy .
i.e 10 units of each product can be produced.
Example 2. A pharmaceutical company manufactures three types of
medicines; M 1, M 2 and M 3. Each type of medicine contains three drugs
D1, D2 and D 3 as given in the following table:
Medicine Drug (in mg)
D1 D2 D3
M1 1 3 2
M2 2 1 3
M3 3 2 1
What amount of the three medicines should be taken to consume 14 mg of
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91 Ans: Let x units of M 1, y units o f M 2 and z units of M 3 be consumed. Then
according to the given information we have the following system of linear
equations:
23 1 432 1 123 1 1xyz
xy z
xy z
123Now , 3 1 2 1(1 6) 2(3 4) 3(9 2) 5 2 21 18231D
14 2 311 1 2 14(1 6) 2(11 22) 3(33 11) 70 22 66 1811 3 1xD
1 14 33 11 2 1(11 22) 14(3 4) 3(33 22) 11 14 33 362 11 1yD
121 43 1 11 1(11 33) 2(33 22) 14(9 2) 22 22 98 542 3 11zD
By Cramer’s rule we have , and ,y xzD DDxy zDD D
18118x ,36218y and 54318z.
1 unit of M 1, 2 units of M 2 and 3 units of M 3.
6.6 LET US SUM UP
In this chapter we have learn:
Determinant of square matrices.
Properties of determinant.
Use of Cramer’s rule for solving system of equations.
6.7 UNIT END EXERCISE
1) Evaluate the following determinants :
(a) 10 563 (b) 100 2013 3 (c) 417263121 (d) 10 1 413 2401 munotes.in
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92 (e) xyxyxyxy (f) 0abcbaca b
2) Solve the following to find the value of x.
(a) 23511x (b) 23212x
x (c)31502x
x
(d) 10 312 002 1x
x (e) 21 14201 1xx
x
3) Evaluate the following determinants without expanding actually.
1. 111222345 2. 111122133 3. 22
22
22yz y z y zz x zx z xxy x y x y
4) Without expanding actually prove the following :
1. 3 6 112 7 11 202 6 12 2.
222111() () () ( )abc a b b c c a a b cabc
3. 4ab b ab b c c abcac a c
4. 4xxyxzzxyzx x y zyyxzx
5. 11 ( )( )( )1xy x yyz y z x y y z z xxz z x 6. 1101abc
bac
cab
5) Solve the following system of linear equations using Cramer’s rule.
1.251xy, 38 2xy. 2.36xy, 2 5 12xy
3.2xy, 2 3 14xy 4.21 ,xy32xy
5.32 5 ,xy33xy 6.7 10xy, 31 4xy.
7. 7 13xyz, 32 6xy z , 42 1xy z .
8.52xy z , 32 3xy z , 4233xyz
9.3 8 10xyz , 34 0xy z , 2 5 6 13xyz
10.26xyz, 3xyz , 32 7xy z
11.0.2 0.3 0.1 0.9xyz,0.1 0.1 0.1 0.9xyz, 0.3 0.1 0.1 0.1xyz munotes.in
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Determinant
93 12.38 1 0xyz , 34 0xy z , 2561 3xyz
13. 2342 9xyz, 238xyz, 3 2 17xy z
14.527xz, 232xyz, 32 6xy z
15.64 1xy z , 323 1xyz , 52xy z
Word Problems:
6) Mr. Manish invests amount of Rs. 20,000 in three schemes at the rate
of 8%, 10% and 12 % p.a. He had a combined income of Rs. 2,060. The
total income from the second and third scheme was Rs. 1,660. Find the
amount Mr. Manish invested in each scheme using Cramer’s rule.
7)Three students Shahnawaz, Shoaib and Sakina invested their savings of
Rs. 750, Rs. 1200 and Rs. 1545 in 3 differ ent shares with prices x, y and z
respectively as follows. Find the price of each share using Cramer’s rule.
Student share 1 share 2 share 3
(x) (y) (z)
Shahnawaz 6 20 30
Shoaib 30 50 20
Sakina 42 40 43
8) Dr. Ghate ‘s earnings (E) from his clinic is a linear function of a fixed
expense of maintenance (M) plus the number o f patients(P), i.e.E = xP –
M. If his first half yearly earning is Rs.60, 000 with P = 1440 and the
second half yearly earning is Rs. 80,000 with P = 1840, find the fixed
maintenance per month and the average charge per patient using Cramer’s
rule.
9) The prices per unit of three commodities X, Y and Z are Rs. x, Rs. y
and Rs. z respectively. A person ‘ A’ purchases 3 units of X, 5 units of Y
and sells 4 units of Z. ‘ B’ purchases 2 units of X, 1 unit of Z and sells 3
units of Y. ‘ C’ purchases 4 units of Y, 6 units of Z and sells 1 unit of X. In
this transaction A, B and C earn Rs. 6000, Rs. 5000 and Rs. 13000
respectively. Find the prices per unit of the three commodities.
10) Three friends Kanishka, Neha and Darshana invested their pocket
money of Rs. 1 450, Rs. 1800 and Rs. 950 in buying three items A, B and
C from the market. Kanishka buys 7 units of A, 9 units of B and 10 units
of C. Neha buys 6 units of A, 12 units of B and 20 units of C. Darshana
buys 4 units of A 5 units of B and 10 units of C. For mulate this problem
into system of linear equations and solve using Cramer’s rule to give the
price per unit of each item A, B and C.
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94 11) A mixture is to be made of three foods X, Y and Z. Each type consists
of nutrients A, B and C as shown below :
mg per gram of
Food Nutrient A Nutrient B Nutrient C
X 3 3 2
Y 2 1 2
Z 4 2 0
How to form a mixture containing 8 mg of A, 5 mg of B and 6 mg of C?
12) The quarterly sales of a s alesman for three products P 1, P2 and P 3 of a
company are given below with the total commission at different rates:
Quarter P1 P2 P3 Commission
I 100 80 40 600
II 120 100 80 860
III 50 80 100 740
Find, using Cr amer’s rule, the rate of commission on sale per unit of the
products P 1, P2 and P 3.
13) For the following two related market models (a) 52sa bQP P and
4da bQP P (b) 53sa bQP P and 14 2 3da bQP P ; find the
equilibrium prices for each market model by Cramer’s rule.
(Hint: Equilibrium conditions are :sdQQfor both models.)
14) For the following two related market models (a) 15 2sa bQP P and 38 2dbQP (b) 53 2sa bQP P and 11 7 4da bQP P ; find the
equilibrium prices for each market model by Cramer’s rule.
15) The equilibrium conditions for three market models is given below:
1235 6 4 15PP P, 1237 4 3 19PPPand 12 32 6 46PP P . Find the
equilibrium price for each market.
6.8 LIST OF REFERENCES:
Business Mathematics by QaziZameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Busine ss Mathematics by S.R. Arora, Taxmann
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95 7
INVERSE OF MATRIX AND
APPLICATION
Unit Structure
7.0 Objective
7.1 Introduction
7.2 Inverse of a Matrix
7.3 Inverse of a matrix using elementary transformation:
7.4 Inverse of a matrix using Adjoint method:
7.5 Application of matrix inversion to input -output analysis
7.6 Let us sum up
7.7 Unit end exercise
7.8 List of References
7.0 OBJECTIVE:
After going through this chapter you will able to know:
Define inverse of matrix.
Elementary transformation method to find inverse of matrix.
Adjoint method to find inverse of matrix.
Application of inverse of matrix.
7.1 INTRODUCTION:
In the previous chapter we have seen how to add, subtract and multiply
matrices. Though there is no division of two matrices as such, for a square
matrix A, 1/A is written as A-1 and is called as inverse of A. The concept
of inverse of a matrix is useful in solving system of linear equations,
input -output analysis etc.
A very important fact regarding this is that, not every matrix has an
inverse. Only those matrices whose determinant i s not zero have a inverse.
7.2 INVERSE OF A MATRIX:
This leads us to define two more types of matrices :
Singular Matrices : A square matrix X is said to be singular matrix if its
determinant is zero, i.e. 0X. munotes.in
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96 Non-Singular Matrices : A square matrix X is said to be non -singular
matrix if
its determinant is non -zero, i.e. 0X.
A matrix is said to be invertible if it has an inverse.
Before moving ahead let us remember the following properties.
Every square matrix is not invertible.
A square matrix is invertible iff it is non -singular.
Inverse of a matrix is unique.
For a square matrix A of order n, if A-1 is its inverse then A.A-1 = I =
A-1.A, where I is the identity matrix of order n.
The inverse of the inverse of an invertible matrix is itself,
i.e.11()AA
The inverse of transpose of a matrix is transpose of its inverse,
i.e.11() ( )TTAA.
If ,AB are two invertible matrices of same order then, 11 1() .AB B A
Note that the RHS is11.BA and not 11AB, as we know that matrix
multiplication is not Commutative.
There are many methods to find inverse of an invertible square matrix. In
this section we are going to learn to find inverses using elementary row(or
column ) transformations.
7.3 INVERSE OF A MATRIX USING ELEMENTARY
TRANSFORMATION:
Elementary Operations : An elementary operation is either an operation
on row or on column of any of the following three typ es:
Type 1: Interchanging of two rows (or columns). e.g. 13RR, 23CC
Type 2: Scalar multiplication of any row (or column). e.g. 23R, 114C
Type 3: The addition of multip le of any row(or column) to another row
(or
column). e.g. R 1 + 2R 2, 123CC.
To find inverse of a matrix, we follow the following steps :
STEP I : We first check that the given matrix is non -singular, i.e. its
determinant is non -zero.
STEP II : Then a partitioned matrix is created by putting an identity
matrix next to the given matrix separated by a vertical line. munotes.in
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97 STEP III : Now using the row operations the left of the partition is made
an identity matrix proceeding columnwise . Thus what we get on the right
of the partition is the required inverse of the given matrix.
Let us now find inverse of some matrices using the above steps.
Find the inverse of the following matrices, if they exists:
(i) 3162 (ii) 5113 (iii) 49 547 412 1 (iv) 71 433 082 2
Ans : ( i) Let A = 3162
3166062A
Since |A|=0, the given matrix is singular and hence not invertible.
i.e. A-1 does not exists.
(ii) Let B = 51135115 1 16 013B
B is non -singular, B – 1 exists.
Now consider the partitioned matrix 51 1 013 0 1
12RRgives, 13 0 151 1 0 ..(1)
215RRgives 13 0 10 16 1 5 ..(2)
21
16Rgives 1 3 0 1150 1 16 16 ..(3)
123RRgives 311 0 16 16150 1 16 16 ..(4)
Thus the inverse of the given matrix is 13116 161516 16B munotes.in
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98 or 13111516B.
Let us understand the solution stepwise:
In the first step we checked the determinant is non -zero and hence
proceeded ahead by creating a partitioned matrix.
Then we transform left side of the partition to an identity ma trix by
making the first column as10. To make the first entry as 1 we use the
first type of elementary operation of interchanging two rows (as the
given matrix already had 1 in the second row), as shown in (1).
Then we use 215RR to make the second entry in the first column
as 0.
Now we make the second column as 01. To make the second entry 1
we use second type of elementary operation21
16R.
Finally to mak e the first entry of second column as 0 we use the
operation123RR.
Simultaneously same operations are performed on the right side of the
partition. The right side of the partitioned matrix in the last step is our
required inverse.
When we are using row operations the left side matrix should be
transformed to identity matrix column -wise as described above.
(iii) Let C
=49 547 412 149 5| | 4 7 4 4(7 8) ( 9)(4 4) 5( 8 7) 4 0 5 112 1C
0CQ , C-1 exists.
Now consider the partiti oned matrix 49 5 1 0 047 4 0 1 012 1 0 0 1
13RRgives 12 1 0 0 147 4 0 1 049 5 1 0 0 munotes.in
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99 21 314 and 4RR RR gives 12 1 0 0 101 0 0 1 401 1 1 0 4
12 3 22 and RR R Rgives10 1 0 2 701 0 0 1 400 1 1 1 8
13RRgives100 1 11010 0 1 4001 1 1 8
111 1014118C.
(iv) Let D = 71 433 082 271 43 3 0 7( 6) 1(6) 4(6 24) 42 6 72 36 082 2D 0DQ D – 1 exists.
Consider 71 4 1 0 033 0 0 1 082 2 0 0 1
13RRgives 11 2 1 0 133 0 0 1 082 2 0 0 1
21 313 and 8RR RR gives11 2 1 0 100 631306 1 8 8 0 7
31
6Rand21
6R gives 11 2 1 0 100 1 1 / 21 / 61 / 201 3 4 / 3 0 7 / 6
23RRgives11 2 1 0 101 3 4 / 3 0 7 / 600 1 1 / 21 / 61 / 2 munotes.in
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100 12RRgives10 11 / 3 0 1 / 601 3 4 / 3 0 7 / 600 1 1 / 2 1 / 6 1 / 2
13 2 3 and 3RR R R gives100 1 / 61 / 61 / 3010 1 / 61 / 21 / 3001 1 / 21 / 61 / 2
11/6 1/6 1/3 1 1 211/6 1/2 1/3 1 3 261/2 1/6 1/2 3 1 3D
7.4 INVERSE OF A MATRIX USING ADJOINT
METHOD:
The inverse of a matrix A is calculated by the formula:
5.6.1 Adjoint of a 2 x 2 matrix:
Let A = abcd then Adjoint of A = adj( A) = dcba
And
Example 18: Find the inverse of 2371using adjoint method.
Ans:2371A= 2 + 21 = 23
Now, adj( A) =
Hence
5.6.2 Adjoint of a 3 x 3 matrix:
Let A = 11 12 1321 22 2331 32 33aaaaaaaaathen A– 1 = 1Adcda munotes.in
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101 Matrix of Co -factors : C = 11 12 1321 22 2331 32 33AA AAA AAA A
Where each Aij : determinant of the minor matrix of A deleting the ith row
and jth column.
Observe the signs of the cofactors in side the matrix. Th ey are alternately
+and – .
Example 19: Find the inverse of A =12 141 863 5using adjoint method.
Ans:12 141 863 5A = 1(– 29) – 2 (–28) – 1(18) = 9
Now we find all the nine cofactors of the matrix A as follows:
A11 is obtained by deleting the 1st row and 1st column of A
A11 = 1835 = – 29
A12 is obtained by deleting the 1st row and 2nd column of A
A12 = 4865= – 28
A13 is obtained by deleting the 1st row and 3rd column of A
A13 = 4163= 18
A21 is obtained by deleting the 2nd row and 1st column of A
A21 = 2135= 13
A22 is obtained by deleting the 2nd row and 2nd column of A
A22 = 1165= 11
Adj(A) = CT munotes.in
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102 A23 is obtained by deleting the 2nd row and 3rd column of A
A23 = 1263= – 9
A31 is obtained by deleting the 3rd row and 1st column of A
A31 = 2118= 15
A32 is obtained by deleting the 3rd row and 2nd column of A
A32 = 1148= 12
A33 is obtained by deleting the 3rd row and 3rd column of A
A33 = 1241= – 9
Thus, the matrix of cofactors: C =11 12 1321 22 2331 32 33AA AAA AAA A= 29 28 1813 11 915 12 9
Adj(A) = CT = 29 13 1528 11 1218 9 9
Hence A– 1 = 1Aadj(A) = 1929 13 1528 11 1218 9 9
Inversion method:
Example: Solve the following equations using Matrix inversion method
Solution: Given equations can be written as
The corresponding matrix equation AX=B for m is
Where A=
Let
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103 Multiple by
on both sides we get
----I
Now
A=
⟹
Hence
is exists.
Adjoint of
Using equation I we get
Hence,
Example:Solve the following equations using Matrix inversion method
Solution: Given equations can be written as
The corresponding matrix equation AX=B form is
Where A=
Let
Multiple by
on both s ides we get
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104
----I
Now
A=
⟹
Hence
is exists.
A=
First find cofactor of A:
Cofactor of A =
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105 Adjoint of
Using equation I we get
Hence,
7.5 APPLICATION OF MATRIX INVERSIO N TO
INPUT -OUTPUT ANALYSIS
Input -Output analysis is a technique which was put forward by Wassily
W. Leontief. It is a mathematical method which uses matrix technology to
study the interdependencies of different sectors on each other in a
economy. The outpu t of one industry may form the input of another
industry and vice versa. The effect of changes in the input/output of
various sectors is calculated using matrix inversion method.
Let us now first go through the terminology and concepts of input -output
analysis which is done using the input -output table:
Input : The material demanded by a producer for his production is called
as input. It is the expenditure occurred to the producer.
Output : The outcome of a production process is called as output. It is the
income gained by the producer.
Total cost : The total money value of all the inputs is called as the total
cost of the producer.
Total revenue : The total money value of all the outputs is called as the
total revenue of the producer.
Input – Output table : This is a matrix consisting of various rows and
columns representing the distribution of the outputs and inputs of different
sectors in an economy.
Now, let us see an example of an input -output table consisting of three
interdependent sectors.
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106 Producer
sector
(output) Consumer sector
(input) Final
demand Total
output
I II III
I a11 a12 a13 d1 x1
II a21 a22 a23 d2 x2
III a13 a32 a33 d3 x3
Table 5.1
Recall that in the expression aij, the first suffix letter ‘ i’ represents the row
and the second letter ‘ j’ represents the corresponding column.
a11 is the money value of the output sector I which is used by the input
sector I. d1 is the money value of the demand of the output and x1
represents the money value of the total output of the sector I. Similarly, a23
is the money value of the output sector II which is used by the input sector
III.d2 is the money value of the demand of the output and x2 represents the
money value of the total output of the sector II.
Thus, from the table 5.1 we can write the followi ng equations;
x1 = a11 + a12 + a13 + d1, x2 = a21 + a22 + a23 + d2 and
x3 = a31 + a32 + a33 + d3. … (1)
Technical Coefficients : We know that the output of one sector forms the
input of another sector. The number of units of output of a sector required
to produce one unit output of another sector are called as the technical
coefficients. These coefficients are the ratio of the money value of the
output of one sector required by another sector to the money value of the
total output of the supplying sector .
From the table 5.1 and eqn (1) we can write the technical coefficients as
follows:
b11 =111ax, b12 =122ax, b13 =133ax
b21 =211ax, b22 =222ax, b23 =233ax … (2)
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107 Technology matrix : The matrix of all the technical coefficients is called
as the technology matrix. For the above example, we have the technology
matrix as: B =11 12 1321 22 2331 32 33bbbbbbbbb
From (1) and (2), we have the following equations:
x1 = a11 + a12 + a13 + d1 = b11x1 + b12x2 + b13x3 + d1
x2 = a21 + a122 + a23 + d2 = b21x1 + b22x2 + b23x3 + d2 … (3)
x3 = a31 + a32 + a33 + d3 = b31x1 + b32x2 + b33x3 + d3
If we denote X =123xxx, D =123ddd then the above eqn (3) becomes:
X = BX + D X – BX = D (I – B)X = D
Thus, X = (I – B) – 1D
The importance of the above matrix equa tion is that if the final demand
change then the corresponding change in the total output can be computed
easily as the matrix ( I – B)– 1 is already known and is independent of the
final demands hence is constant. The matrix ( I – B) is called the
Leontief’ s matrix .
Before we proceed in solving problems related to input -output tables let us
go through the assumptions of input -output analysis:
No joint product by different sectors, i.e. every sector produces only
one product.
The ratio of input and output is fixed.
Constant returns of scale.
No external economies or diseconomies.
Prices remain unchanged.
Example 20: From the following input -output table find the technical
coefficients, technology matrix and the total output of the two industries.
Industry X Y Final
demand
X 100 60 240
Y 150 80 70
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108 Ans: The total output for sector X is 100 + 60 + 240 = 400
The total output for sector Y is 150 + 80 +70 = 300.
The total input for sector X is 100 + 150 = 250
The total input for sector Y is 60 + 80 = 140
The above data can be put in the given table as follows:
Now, the four technical coefficients are:
b11 = 111ax= 100400= 0.25, b12 = 122ax= 60400= 0.15
b21 = 211ax= 150300= 0.2, b22 = 222ax= 80300= 0.27 the technology matrix is B = 0.25 0.150.2 0.27
Example 21: From the following input -output table find the technical
coefficients, technology matrix and the total output of the tw o industries.
If the final demand changes to 200 and 100 for the sectors X and Y
respectively, find the total output.
Industry X Y Final
demand
X 20 50 130
Y 30 70 50
Ans:
The total output for sector X is 20 + 50 + 130 = 200
The total output for sector Y is 30 + 70 +50 = 150.
The total input for sector X is 20 + 30 = 50
The total input for sector Y is 50 + 70 = 120
Produ cer
sector
(output) Consumer sector
(input) Final
demand Total
output
X Y
X 100 60 240 400
Y 150 80 70 300
Total
input 250 140 310 700
Producer
sector
(output) Consumer sector
(input) Final
demand Total
output
X Y
X 20 50 130 200
Y 30 70 50 150
Total
input 50 120 180 350 munotes.in
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109 The above data can be put in the given table as follows:
Now, the four technical coefficients are:
b11 = 111ax= 20200= 0.1, b12 = 122ax= 50200= 0.25
b21 = 211ax= 30150= 0.2, b22 = 222ax= 70150= 0.47 the technology matrix is B = 0.1 0.250.2 0.47
To find the total output ( X) we will u se the formula: X = (I – B) – 1D
Now, B =0.1 0.250.2 0.47, D= 200100, X = 12xxand I = 1001 (I – B) = 1 0 0.1 0.250 1 0.2 0.47 = 0.9 0.250.2 0.53
Using the formula to find inverse of a 2 x 2 matrix by adjoint method from
section 5.6.1 , we have
Adj(I – B) = 0.53 0.250.2 0.9and IB= 0.9 0.250.2 0.53
= 0.427 (I – B) – 1 = 10.4270.53 0.250.2 0.9= 1.24 0.590.47 2.11 The total output is given by X = (I – B) – 1D = 1.24 0.59 2000.47 2.11100 X = 1.24 x 200 0.59 x 1000.47 x 200 2.11 x 100= 307305 If the final demand changes to 200 and 100 the corresponding total
output of the two sectors X and Y are:
Total output of sector X is x1 = 307
The total output of sector Y is x2 = 305
Example 22: The following table gives the technology matrix of three
sectors of an economy. If the final demands change to 100, 150 and 50
find the total output using Leontif’s matrix.
Sector P Q R
P 0.2 0.34 0.15
Q 0.18 0.67 0.5
R 0.45 0.28 0.75 munotes.in
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110 Ans: Let the technology matrix be B = 0.2 0.34 0.150.18 0.67 0.50.45 0.28 0.75
(I – B) = 1 0 0 0.2 0.34 0.150 1 0 0.18 0.67 0.50 0 1 0.45 0.28 0.75 = 0.8 0.34 0.150.18 0.33 0.50.45 0.28 0.25 IB= 0.8 0.34 0.150.18 0.33 0.50.45 0.28 0.25
= 0.8( –0.0575) + 0.34( –0.27) –0.15(0.1989) IB= – 0.11
The matrix of cofactors = C = 0.0575 0.27 0.19890.043. 0.1325 0.3770.2195 0.427 0.2028
Adj(I – B) = CT = 0.0575 0.043 0.21950.27 0.1325 0.4270.1989 0.377 0.2028 (I – B)– 1 = 1IBadj(I – B) = 10.110.0575 0.043 0.21950.27 0.1325 0.4270.1989 0.377 0.2028 (I – B)– 1 = 0.52 0.39 22.45 1.2 3.881.81 3.43 1.84
Now D= 10015050, thus, the change in total output is given by the formula
X = (I – B)– 1D = 0.52 0.39 22.45 1.2 3.881.81 3.43 1.8410015050= 10.5259241.5
Thus, the changes in final demand to 100, 150 and 50 lead to the following
total output of the three sectors:
For sector P: x1 = 10.5, for sector Q: x2 = 259 and for sector R: x3 = 241.5
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111 7.6 LET US SUM UP:
In this chapter we have learn:
Definition inverse of matrix. And its properties.
Different method of calculation of inverse of matrix.
Different ap plication of inverse of matrices.
7.7 UNIT END EXERCISE:
Q.1 Find the inverse of the following matrices using elementary row
operations and also by adjoint method:
1. 3411 2.6253 3.7121 0 4.3612
5.112132313 6. 11 132 423 5 7. 11 311 112 1 8.121212003
9. 12 443 210 3 10. 32613 121 0
Q.2 Solve the following system of equations by inversi on method.
i)
ii)
iii)
iv)
v)
Q.3From the following input -output table find the technical coefficients,
technology matrix and the total output of the two industries, find the total
output with respect change in demand.
i)
Industry X Y Final
demand Demand change
X 40 50 120 250
Y 50 70 70 100
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112 ii)
Industry X Y Final
demand Demand change
X 20 40 80 120
Y 30 50 100 150
iii)
Industry X Y Final
demand Demand change
X 40 60 150 300
Y 50 80 100 200
iv)
Industry X Y Final
demand Demand change
X 30 70 180 280
Y 60 90 130 220
v)
Industry X Y Final
demand Demand change
X 20 35 150 180
Y 25 60 130 160
7.8 LIST OF REFERENCES
Business Mathematics by QaziZameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
munotes.in
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113 8
DERIVATIVES
Unit Structure
8.0 Objective
8.1 Introduction
8.2 Derivativesas a measure of rate
8.3 Rules of Derivatives
8.4 Second order derivatives
8.5 Let us sum up
8.6 Unit end exercise
8.7 List of References
8.0 OBJECTIVE
After going through t his chapter you will able to know:
Definition of Derivatives.
Standard formula of Derivatives.
Rule of Derivatives.
8.1 INTRODUCTION
In business and economics many decisions are made on the basis of
marginal analysis, like marginal cost, marginal revenue o r marginal
productivity etc.
The branch of mathematics which deals with such type of marginal
analysis, or in other words, the process wherein rate of change in one
variable with reference to a change in other variable is to be analyzed is
called Calculus . The study of Calculus is divided onto two major aspects:
Differential Calculus and Integral Calculus . In this chapter we will be
studying the former i.e. differential calculus.
Before we begin with differentiation, let us first understand limit of a
funct ion.
Limit of Functions: Before introducing the concept of derivative of a
function it is important to understand the concept of limit of a function.
We would not define limit of a function formally, as it is not our area of
interest. But an overview of thi s concept is given below.
Let y be a function of x defined as: 242xyx munotes.in
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114 If we want to know the value of y when x = 2, one would simply put x = 2
in above equation.
This would give us, 224 022 0y which is a value not defin ed in
mathematics!
To solve this problem we have to understand, what are the values of y
when x approaches the value 2. The following tables give a clearer picture:
Table 1 Table 2
X y x y
0 2 4 6
1 3 3 5
1.5 3.5 2.5 4.5
1.8 3.8 2.1 4.1
1.9 3.9 2.01 4.01
1.99 3.99 2.001 4.001
0 1 2 3 4
The Table 1 gives the value of y as x approaches the value 2 from left side
of the number line i..e. from 0 to 2, while the Table 2 gives the valu e of y
as x approaches the value 2 from right side of the number line i.e. from 4
to 2. As we can see from both the tables that as xnears the value 2, the
value of y becomes approximately 4. This can be said as follows: the limit
of y as x tends to (approa ches to) 2 is 4, which is represented symbolically
as2lim 4xy.
In general if()yf x, then we say that the limit of y (if it exists) is l as x
tends to a, if y approaches the value l as x approaches a, from both sides of
the number line. Symbolically this is written aslim lim ( )xa xayf x l.
This is an informal definition of limit of a function. A more mathematical
definition can be obtained by curious students from any book on Calculus.
8.2 DERIVATIVESAS A MEASURE OF RATE
Differentiation is the process of finding the derivative of a function. As
mentioned in the introductory part above, economics and problems in
business deal with marginal analysis of variables. So we are interested in,
what and how much is the change in y with reference to a change in x,
where()yf xis a continuous function of x.
Let δ x be the small change in x and δ y be the corresponding change in y. munotes.in
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115 Then we have y + δy = f(x + δx) δy = f(x + δx) – y
Dividing both sides by δ x, we get () ( )yf x xf xxx
Taking limit as0x, on both sides, we get
0 0() ( )lim lim
x xyf x x f xxx
The limit on the LHS if exists, is called the derivative of ywith respect to
(hereafter written as w.r.t.) x and is denoted as dydx.
Thus, dydx is the rate of change of y w.r.t. the change in x and is called
as the derivative of y w.r.t. x
Remark :
1. The expressiondydx, should not be misinterpre ted as two values dy
divided by dx
2.ddxis the differential operator, which when operated on functions gives
its derivative. For example, consider our demand function()Df P. Here
P is the independent variable, so the derivative of D w.r.t. P is ()dDdP or
in short dDdP
3. One more important thing to remember is that the variables (both
dependent and independent) may keep changing in the expression of the
derivative from prob lem to problem, i.e.dydx is not the only notation for
derivative it can be dPdD or dQdPordCdx as the function may be.
4. If dydxexists for a funct ion y = f(x) then we say that y is a differentiable
function of x.
Let us now study derivatives of some standard functions.
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116 1. Derivative of a constant function
Let y = k be a constant function, where k is any real value then () 0dy dkdx dx
Thus, Derivative of a constant function is zero
All real values like 11, -304, 5/17, -4.234, log3, e, 2etc are examples of
constant functions.
Solved examples:
Find the derivatives of the following functions:
(i) 12, (ii) -10, (iii) 1/13, (iv) -7/16, (v) 0, (vi) log 4,
(vii) 2
Ans : Since all the functions above are constant functions, their derivative
is zero using the above formula .
i.e. (12) ( 10) (1/13) ( 7 /16) (0) (log 4) ( 2)0dd d d d d ddx dx dx dx dx dx dx
2. Derivative of a power function
If y = xn is a power function of x, where n is any real number then, 1.ndynxdx
Solved examples :
Find the derivatives of the following functions:
(i) x 12, (ii) x -3, (iii) 2/3x, (iv) 6/7x, (v) 1/2x, (vi) x, (vii) 1x
(i) 12 12 1 1112 12.dxxxdx
(ii) 33 1 4(3 ) . 3dxx xdx
(iii) 22 2 3 1133 33222...333dxxx xdx
(iv) 6 6 6 7 13177 77666...77 7dxxx xdx
(v) 111 2 1122 22
1/2111 1 1...222 2 2dxxxxdx xx
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117 1/2xxQ , from above problem, 12dxdxx
(vi) 1/21/211()dd dxdx dx dx x x
=112 1
2x
= 1221
2x
= 321
2x
3/212dx
dx x
3. Derivative of an expo nential function
If y = ax is an exponential function of x, where a> 0, then logx dyaadx
Solved examples :
Find the derivatives of the following functions: 2x, 9 x
(i) Let y = 2x
(2 ) 2 log2xx dy ddx dx
(ii) Let y = 9 x
(9 ) 9 log9xx dy ddx dx
4. If y = ex, then by the above formula:
( ) logxx xdy dee e edx dx (since log 1e)
Thus, we have () dxxeedx
5. Derivative of a logarithmic function
Iflogyx, (where x ≠ 0) then 1(log )dy dxdx dx x
Hence, (log )1dx
dxx
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118 Let us sum up all the formulae of differentiation in tabular form:
Function : y Derivative : dydx
1. k, constant function () 0dkdx
2. xn 1() .nndxn xdx
3. ax, (where a> 0) ( ) logxx daa adx
4. e x ()xxdeedx
5. log x (where x ≠ 0 ) 1(log )dxdx x
Here are some problems for you to try now !
8.3 RULES OF DERIVATIVES
Now when we are through with the basic formulas of differentiation, one
would be interested in knowing the derivatives of 45x, 7logx, loglog6x, 2 3 4logxeex x or 2 .3xxetc. F or that we require certain rules of
differentiation. In this section we will study these rule of differentiation .
Rule I : Derivative of scalar multiple of a funcDion :
I ()fx is a function of x, then
() [ () ]ddkf x k f xdx dx , where k is a scalar ( i.e. constant)
Solved problems based on Rule I :
Differentiate the following functions W.r.t. x :
(i) 34x, (ii) 1/33x
, (iii) 2x, (iv) 3.4x, (v)3log3x, (fi) 2log3x
Ans:
(i) Let 34yx
Differentiating y w.r.t. x, we have, 33() ( 4 ) 4 ( )dd dyx xdx dx dx31 24(3. ) 12xx munotes.in
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119 212dyxdx .
(ii) Let 1/33xy
Differentiating y w.r.t. x, we have, 11/311/3 311 1() ( ) . .33 3 3dd x dyx xdx dx dx (Here the scalar is 13) 4/31
9dyxdx
(iii)Let 2yx.
Differentiating y w.r.t. 119x, we Have,
1( ) 2 2 ( ) 2.
2dd dyx xdx dx dxx 1dy
dxx .
(iv Let 3.4xy
Differentiating y w.r.t. x, we have,
( ) 3.4 3 (4 )xxdd dydx dx dx 3(4 log 4)x dy
dx
(v) Let 3log3xy
Differentiating y w.r.t. x, we have,
31() ( 3 )log3 log3xxdd dydx dx dx (Here the scalar is 1log3)
3 log33log3xxdy
dx
(vi) Let2log3xy
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120 Differentiating y w.r.t. x, we have,
2log 2( ) (log )33ddx dyxdx dx dx (Here the scalar is 23) 21 2.33dydx x x
Rule II : Derivative o f sum or difference of two (or more) functions
If f(x) and g(x) are functions of x, then the derivative of () ()fx g x is
given by: [() () ] [() ] [ () ]dd dfx g x fx g xdx dx dx (Sum Rule )
And the derivative of () ()fx g x is given by [() () ] [() ] [ () ]dd dfx g x fx g xdx dx dx (Difference Rule )
This rule can also be extended to more than two functions as given below: [( ) ( ) ( ) ] [( ) ] [( ) ] [ ( ) ]dd d dfx g x h x fx g x h xdx dx dx dx
Solved problems based on Rule II :
Differentiate the following functions w.r.t x :
(i) log5xe, (ii) 2 logxx, (iii) 33xx, (iv) log 7xx
Ans:
(i) Let y = log5xe
Differentiating y w.r.t. x, we have, ( ) ( log5) ( ) (log5) 0xx x dd d dye e edx dx dx dx
(l o g 5 )xxdeedx.
(ii) Let y = 2 logxx
Differentiating y w.r.t. x, we have, ( ) (2 log ) (2 ) (log )xx dd d dyx xdx dx dx dx 1(2 log ) 2 log 2xxdxdx x . munotes.in
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121 (iii) Let 33xyx
Differentiating y w.r.t. x, we have,
33 3 1( ) ( 3 ) ( ) (3 ) 3 3 log3xx x dd d dyx x xdx dx dx dx
32( 3 ) 3 3 log3xx dxxdx .
(iv) Lety = log 7xx
Differentiating y w.r.t. x, we have, ( ) (log 7 ) (log ) (7 )xxdd d dyx xdx dx dx dx
3 1(3 ) 7 l o g 7xx dxdx x .
Now get ready for some problems for practice!
Rule III : Derivative of product of two functions:
If f(x) and g(x) are functions of x, then the derivative of product of f(x) and
g(x) i.e. f(x).g(x) is given by:
() .() () [() ] () [() ]dd dfxg x g x fx fx g xdx dx dx (Product Rule )
(This can be remembered as 2nd into derivative of 1st+ 1st into derivative
of 2nd!)
Solved problems based on Rule III :
Differ entiate the following functions w.r.t x :
(i) 2 .3xx, (ii) 2logxx,
Ans:
(i) Let y = 2 .3xx
Differentiating y w.r.t. x, we have, ( ) (2 .3 ) 3 (2 ) 2 (3 ) 3 (2 log 2) 2 (3 log3)xx x x x x x x x x dd d dydx dx dx dx (2 .3 ) 2 .3 (log 2 log3)xx xx d
dx .
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122 (ii) Let y = 2logxx
Differentiating y w.r.t. x, we have,
22 2 2 1 21() ( l o g) l o g ( ) ( l o g) l o g( 2 )dd d dyx x x x x x x x xdx dx dx dx x
2( log ) 2 logdxxx x xdx
Rule IV : Derivative of quotient of two functions :
If f(x) and g(x) are functions of x, then the derivative of quotient of f(x)
and g(x), i.e. ()()fxgx is given by
2( ) [ ( )] ( ) [ ( )]()
() [() ]ddgx f x f x gxdf x dx dx
dx g x gx (Quotient Rule )
Solved problems based on Rule IV :
Differentiate the following functions w.r.t x :
(i) 43xx (ii) 4logxx
Ans:
(i) Let y = 43xx
Differentiating y w.r.t. x, we have,
223 (4 ) 4 (3 )4 3 4 log 4 4 3 log3()
3 (3 ) (3 )xx xxxx x x x
xx xdd
dd dx dxydx dx
23 4 (log 4 log3) 4 (log 4 log3)(3 ) 3xx xxxdy
dx
(ii) Let y = 4logxx
Differentiating y w.r.t. x, we have,
3
224444 log ( ) (log )(log )4 (1 )log (log ) (log )ddxxx xdy d x x x x xdx dx
dx dx x xx
23(4log 1) 4 (log 4 log 3)(log ) 3x
xdy x x
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123 The act ual problems of differentiation are not segregated like those solved
using rule I, or rule II or so, but would require more than one rule to be
used. In such cases the method to differentiate the given functions is to use
the above rules one by one and sim plify the problem to basic functions
whose derivatives we know. In this section we shall learn how to solve
problems using all the rules and then we shall have some problems for
practice.
Solved problems using all the four rules and formulae :
Differentiat e the following functions w.r.t. x.
(i) 236 l o gxx (ii) 3.log 3xxe (iii) 1/ 44logxx (iv) 2exxxee
(v) 1/33 4.225 l o gxxxx
Ans:
(i) Let y = 236 l o gxx
Differentia ting w.r.t. x, we have,
2 2( ) (3 6log ) (3 ) (6log )dd d dyxxx xdx dx dx dx (usingRule II )
23 ( ) 6 (log )dy d dxxdx dx dx (using Rule I )
21 1 63(2 ) 6 6 6( 1/ )xdyxx x xdx x . (using formulae )
(ii) Let 3.log 3xxey
Differentiating w.r.t. x, we have,
3. 1(3 . )log3 log3xxxxdy d e dedx dx dx (using Rule I )
1(3 ) 3 ( )log3xx xxdy d deedx dx dx (using Rule III )
1(3 log3) 3 ( )log3xx xxdyeedx (using formulae ) 3 (log3 1)log3xxdy e
dx
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124 Differentiating w.r.t. x, we have,
1/4 1/444log logdy d x d xdx dx x dx x (using Rule I )
1/4 1/4
2(log ) ( ) (log )4
(log )ddxx x xdy dx dx
dx x (using Rule IV )
1/4 1 1/4
241 1(log ) ( ) .4 (log )dyxx xdx xx (using formulae )
3/4 3/41/4 1 3/4224 ( )(log ) 4 ( )(log )
44 (log ) (log )dy x x x xxxdx xx
3/441log logdy xdx x x .
(iv) Let 2exxyx ee
Differentiating w.r.t. x, we have,
2( 2 )eexxdy d x d d xxe xedx dx e dx dx e (using Rule II )
12( ) .( )xedy d dxe xdx dx e dx (using Rule I ) 112 . ( ) ( ) .( )xx e dy d dex x e e xdx dx dx e (using Rule III )
11112 .(1 ) ( ) 2( )xx e x x edye x x e x e xe xdx
(using formulae and 01xQ )
(v) Let 1/33 4.22 6logxxyxx
1/3 1/31/3
2(2 6log ) (3 4.2 ) (3 4.2 ) (2 6log )3 4.2
2 6log (2 6log )xxxddxx x x xxdy d x dx dx
dx dx xx xx
(using Rule IV )
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125 1/3 1/3
2(2 6log ) (3 ) (4.2 ) (3 4.2 ) (2 ) (6log )(2 6log )xx dd d dxxx x x xdy dx dx dx dx
dx xx
(using Rule II )
1/3 1 1/3
212 6(2 6log ) 3. ( ) 4(2 log 2) (3 4.2 )3 2
(2 6log )xxxx x xxdy x
dx xx
(using Rule I and formulae )
2/3 1/3
216(2 6log ) ( ) 4(2 log 2) (3 4.2 )
(2 6log )xxxx x xxdy x
dx xx
.
8.4 SECOND ORDER DERIVATIVES
Let y = f(x) be a differentiable function of x. dydxis called the first
derivative of f(x). dydxis itself a function of x and can be differentiated
again w.r.t. x.
If the derivative of dydxexists it is called the second derivative of f(x) and is
denoted as 22dydx. In this way we can proceed to get further derivatives of
higher order.
Thus, for a function y = f(x) dydxis called the first derivative of ()fx 22dyd
dxdxdy
dx
is called the second derivative of f( x)
Before we solve problems related to second derivative of functions,
readers should get themselves thorough with all the formulas and rules of
differentiation.
Solved Examples
1. If236 9yx x , find 22dydx.
Ans: Differentiating y w.r.t. x, we have: 66dyxdx munotes.in
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126 Differentiating dydxw.r.t. x, we get, 22dydx= 6.
2. If4
7/661 2 1 04xyxx , find 22dydx
Ans: Differentiating y w.r.t. x, we have: 31 / 667dyxxdxx
Differentiating dydxw.r.t. x, we get, 225 / 6 3 / 227336dyxx x
dx .
3. If2xyx e, prove that 2222( 1 2 )x dyex xdx
Ans: Differentiating y w.r.t. x, we have:
22(2 ) 2xx x xdyex x e x ex edx
Differentiating dydxw.r.t. x, we get
2
2
22( .1 ) 2xx x xdye xe xe x edx
222 2xxxxex ex e x e
2222( 1 2 )x dyex xdx
4. If log2xxxy, find 22dydx
Ans: Differentiating y w.r.t. x, we have
212 log .1 . log (2 log 2)(2 )xx
xxx x x
x dy
dx
22( l o g 1 )( l o g ) ( l o g 2 )(log 1) ( log )(log 2)(2 ) 2x
xxxx x xx x dy
dx
Differentiating dydxw.r.t. x, we get
22
22 0 (log 2)(log .1 . ) [(log 1) ( log )(log 2)]2 log 2(2 )11x
xxx x x xx
dy xx
dx
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127 2
2(log 2)(log 1) [(log 1) ( log )(log 2)]log 221
xxx x xdy x
dx
8.5 LET US SUM UP
In this chapter we have learn:
Formulas of differentiation
Rule of differentiation and its basic problem
8.6 UNIT END EXERCISE
Q.1 Find the derivatives of the following functions :
(i) -100, ( ii) 21/12, (iii) 32, (iv) log 16, (v) x 4 (vi) 10x,
(vii) 45x, (viii) 16x, (ix) x, (x) 5x
Q.2 Differentiate the following w.r.t. x.
(i) 66x, (ii) 42x, (iii) 2/55x, (iv) 1/147x, (v) 4x
(vi) 14.8x, (vii) 9log9x, (viii) 5 .log5x, (ix) 2xe, (x) 5log4x
Q.3 Differentiate the following functions w.r.t x :
(i) logxex, ( ii) 9xex, ( iii) 2xxe, (iv) 1/3 1/2xx, (v) 66xx
(vi)1/3 1/2 1/6xxx , (vii) 6logxx,(viii)4/57 log 2xx ,(ix) 1010xx
(x) 2222xx.
Q.4 Differentiate the following functions w.r.t x :
(i) .xex (ii) 5logxx (iii) 55xx (iv) 3/22xx (v) logxex
(vi) xeex (vii) xeax (viii) 39xx (ix) 3xx (x) 4xx
Q.5 Differentiate the following functions w.r.t x :
(i) logxx (ii) 66xx (iii) 36xx (iv) 2xxe (v) log7xx
Q.6 Differentiate the following w.r.t. x :
(i) 2(4 5)x (ii)4331 0 1 0 0 0xx x (iii) 5 .6 2 6xx xex (iv) 1x munotes.in
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128 (v)59x
xexe
(vi)25(2 )55xx
x(vii)1/8( 6)log78xxx(viii)3x(ix)10logxx
(x) 2log2xxx (xi) 5/410 4
xxxe
(xii) 312 9xx
x (xiii)2734x
x
(xiv) 22xx(xv)9/7 14 814 loglog 21611 2xxxxxe xe
(xvi)(4 7)xxe (xvii)
4/3 3/4
9/8342 l o g100 16xx xx
(xviii)32loglog 2xxx (xix)12x
x
(xx) (2x + 1)( x – 2)
Q.7 Find the second derivative for the following functions:
(i) 23(4 3)(3 4)xx (ii) 4327481 1xxx (iii)3239xxx
(iv) 3xxe(v) 22(3 )( 3)xx(vi) 2logxx (vii) logxex(viii) 9 log33 log9xx
(ix) 33 9)3logxx (x) logxx
8.7 LIST OF REFERENCES
Business Mathematics by QaziZameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
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129 9
APPLICATION OF DERIVATIVES
Unit Structure :
9.0 Objective
9.1 Introduction
9.2 Application of economic functions
9.3 Elasticity
9.4 Maxima and Minima for a function of single variable
9.5 Second Derivative test for Maxima and Minima
9.6 Let us sum up
9.7 Unit end exercise
9.8 List of References
9.0 OBJECTIVES
After going through this chapter you will able to know:
Application of economic function and its used.
Elasticity of demand and supply.
Maxima and minima of single variable or increasing and decreasing
function.
Extreme value of function using second order derivative test.
9.1 INTRODUCTION
We know that the process of differentiation is very useful in solving
problems related to marginal analysis in business and economics. One
more important a rea of application of derivatives in economics is related
to optimization problems like, profit maximization or cost minimization.
The significant points where the profit for a certain product is maximum
or the cost per unit production is minimum can be fo und using
differentiation. In this chapter we are going to learn these different
applications of derivatives to Economics and Business.
9.2 APPLICATION OF ECONOMIC FUNCTIONS
Marginal Cost :
If the total cost function for producing x units of a product is g iven by C(x)
then the marginal cost is the rate of change of total cost w.r.t x. In other munotes.in
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130
words, marginal cost is approximate cost of production per extra unit of
the product.
Marginal Cost is denoted by MC and is calculated by the formula:
MC = ()dCxdx
Marginal Average Cost
We have already defined average cost ( AC) in the previous chapter. The
marginal average cost is the rate of change of average cost ( AC).
It is denoted by MAC and is given by: dMAC ACdx
Marginal revenu e
If the total revenue function is given by R(x), then the marginal revenue is
the rate of change of total revenue w.r.t. the quantity demanded.
Marginal revenue is denoted by MR and is calculated by the formula:
MR = ()dRxdx
Relation be tween marginal cost ( MC) and average cost ( AC)
We know that CACx , differentiating w.r.t. x, we get
2()dxC Cdd C dxACdx dx x x
221.( )xd C d C CMAC Cdx x dx xxx ()dMAC ACdxQ
1dC CMACx dx x 1MAC MC ACx
Thus, 1MAC MC ACx or 1 dAC MC ACdx x
An analysis of the above relation indicates that if MC = AC then
0dACdx.
Ex.1 . If the total cost function of a
producing a product is given by
Which means the slope of AC
is zerowhen MC and AC cut
each other at the minima of
AC as shown in the adjacent
figure.
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131 320.002 0.05 25 1000Cx x x , find ( a) average cost ( AC), (b) marginal
cost ( MC) and ( c) marginal average cost ( MAC ).
Ans: Given 320.002 0.05 25 1000Cx x x
Now, CACx320.002 0.05 25 1000xx x
x
210000.002 0.05 25 AC x xx … (I)
And dCMCdx. Differentiating C w.r.t. x, we get,
320.002 0.05 25 1000dMC x x xdx
20.006 0.1 25MC x x … (II)
Also, ()dMAC ACdx . Differentiating (I) w.r.t x, we get,
210000.002 0.05 25dMAC x xdx x
210000.004 0.05 MAC xx … (III)
Ex.2 . If 32251 5 1 0 0Cx x x is the total cost function, find ( a) AC, (b)
MC, (c) MC when x = 10 and ( d) actual cost of producing 11th unit of the
product.
Ans:
(a) Average Cost
32
22 5 15 100 1002 5 15Cxx xAC x xxx x … (I)
(b) Marginal Cost
32(2 5 15 100)dC dMC x x xdx dx
261 0 1 5MC x x … (II)
(c) when x = 10,
2
106(10) 10(10) 15 600 100 15xMC
10Rs. 515xMC
(d) Actual cost of producing an extra unit i.e. the 11th unit is calculated as: munotes.in
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132 32 32
11 10[2(11) 5(11) 15(11) 100] [2(10) 5(10) 15(10) 100]xxCC
[2662 605 165 100] [2000 500 150 100]
2322 1750 572
Thus, the actual cost of producing the 11th unit is Rs. 572.
Ex.3 . Verify that 1() ( )dAC MC ACdx x , for the cost function given as
231000 120 10 2Cx x x .
Ans:2321000 120 10 2 1000120 10 2Cx x xAC x xxx x … (I)
2
21000 1000() 1 2 0 1 02 1 0 4ddAC x x xdx dx x x … (II)
23 21000 120 10 2 120 20 6dC dMC x x x x xdx dx … (III)
Subtracting (II) from (III), we get,
221000120 20 6 120 10 2MC AC x x x xx
i.e.2100010 4MC AC x xx
Thus, 2
21 1 1000 100010 4 10 4 ( )dMC AC x x x ACx x x dx x
1() ( )dAC MC ACdx x .
Ex.4 . The total cost function of a product is 3210000 100 103xCx x .
Find the ( a) AC, (b) MC, (c) VC, (d) AVC , (e) dMCdx and ( f) no. of units
when MC = AVC .
Ans:
(a) Average Cost
23 210000 100 10 / 3 10000100 103Cx x x xAC xxx x
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133 (b) Marginal Cost
32210000 100 10 100 203dC d xMC x x x xdx dx
(c) Variable cost
()C FC VC xQ
32100 103xVC x x
(d) Average variable cost
23 2100 10 / 3100 103VC x x x xAVC xxx
(e) 2100 20ddMC x xdx dx 20 2x
(f) when MC = AVC , from ( b) and ( d) we have,
22100 20 100 103xxx x
2
210 03xxx 223 0 0xx
(2 30) 0xx 0 or 15xx
Discarding x = 0, we have MC = AVC when x = 15.
Ex.5 . Find AR and MR if the revenue function is given by 210010xR .
Find the MR when x = 2000? Inter pret your answer.
Ans: (a) 10010RxARxx
(b) 2
10010 5dR d x xMRdx dx
(c) when x = 2000;
200020004005xMR
When an additional unit above 2000 units is sold then the revenue
increases by Rs. 400.
Ex.6 . The total revenue function f or a product is given by
2102 240 1000Rx x Find ( a) AR, (b) MR, (c) MR when x = 15, interpret
your answer and ( d) what is the actual revenue from the 16th unit
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134 Ans: ( a) Average Revenue
2102 240 1000 1000102 240Rx xAR xxx x
(b) Marginal Revenue
2102 240 1000 204 240dR dMR x x xdx dx
(c) When x = 15;
15204(15) 240 3060 240 3300xMR
When an additional unit above 15 units is sold then the revenue increases
by Rs. 3300.
(d) The actual revenue from the 16th unit is calculated as follows; 22
16 15[102(16) 240(16) 1000] [102(15) 240(15) 1000]xxRR
[26112 3840 1000] [22950 3600 1000]
30952 27550 3402
Thus, the actual revenue from the extra unit sold is Rs. 3402.
9.3 ELASTICITY
Elasticity of a function y = f(x) is defined as the rate of proportional
change in y per unit proportional change in x.
Price Elasticity of demand
The price elasticity of demand is the proportionate change in quantity
demanded to proportionate change in price. If D = f(p) or p = g(D) is the
price /demand function then the price elasticity of demand is denoted by
ηd (Greek alphabet eta) and is given by:
dp dDD dp
Since the demand curve has a negative slope, η d is negative. Usually we
take the absolute value of η d. Depending upon the values of η d, we have
different types of elasticity’s as follows:
(i) If |ηd|> 1, the demand is elastic . This means that the demand increases
heavily with a small fall in price and decreases heavily with a small
increase in price.
If |ηd|< 1, the demand is inelastic . This means that the demand
increases in a small proportion with a high fall in price and vice versa.
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135
(ii) If |ηd|= 1, the demand is unitary elastic . This means that the
proportionate change in quantity demanded is equal to the
propo rtionate change in price.
(iii)If, |ηd| = 0 the demand is perfectly inelastic . This means a change in
price does not make any change in quantity demanded.
(iv) If, |ηd|B the demand is perfectly elastic . This means the demand
increases infi nitely with a small fall in price.
The value of |ηd| indicates the percentage increase or decrease in demand
due to 1% increase or decrease in price of a commodity.
The price elasticity of demand is useful to any firm in analyzing the
sensitivity of sale s of a certain commodity, to a change in its price.
Necessary commodities exhibit inelastic demand, e.g. salt, water,
electricity etc. The luxury commodities have a high price sensitivity i.e.
they exhibit elastic demand, e.g. cars, diamonds, gold etc.
The following are examples of graphs of linear demand functions
representing the three types of elasticity’s of demand:
1d1d1d
8.5.2 Price Elasticity of Supply
The price elasticit y of demand is the proportionate change in supply of a
commodity to proportionate change in its price. If ()Sf por ()pg Sis the price /demand function then the price elasticity of
demand is denoted by s(Greek alphabet eta) and is given by:
sp dSS dp
Since the supply curve has a po sitive slope, η S is positive.
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136 Relation between Marginal Revenue ( MR) and Average Revenue ( AR)
Let R = pD be the total revenue function, where p is the price and D is the
quantity of demand of a commodity. We know that, AR =p and dRMRdD.
Differentiating R w.r.t. D, we have, ()dR dpDdD dD 1.dp D dpMR p D pdD p dD
Now, dp dDD dp 1111ddMR p AR .
Thus, 11 MR ARd
The above relationship can also be expressed as follows: dARAR MR
An analysis from the above equation indicates that ( i) if the commodity is
perfectly elastic, i.e.d, then10d MR AR. ( ii) If the
commodity is unitary elastic, i.e.1d, then MR = 0 which means that TR
will remain constant irrespective of increase or decrease in the quantity
demanded.
Ex.7 . If the demand function is given by 220pD D , find the elasticity
of demand when D = 5 and interpre t your answer.
Ans: Given 220pD D . Differentiating w.r.t. D, we get,
220 20 2dp dDD D adD dD
Now, 2(20 ) 1
20 2dp dD D DD dp D D
2020 2dDD
When D = 5, 55 201.520 10Dd
Since 1d, the demand function is elastic.
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137 Ex.8 . If AR = 28 and 1.6, find MR.
Ans: We know that 11 MR AR
Substituting the given values, we get
128 1 28(1 0.625)1.6MR 10.5MR
Ex.9 . If the demand funct ion for a certain commodity is given
by220 5pD , find the demand Dwhen 1.
Ans: Given 220 5pD . Differentiating w.r.t. D, we get
220 5 10dp dDDdD dD
Now, 2(20 5 ) 110p dD DD dp D D 2220 5
10DD
If 1, we have 2
220 5110D
D
2220 5 10DD
252 0D 24D
2D (since 2D)
Ex.10 Mr. Bharat Shah manufactures memory cards. The total revenue
function is given as 20.4 1400Rx x . Find dand interpret it when x is
400, 482 and 560.
Ans: Given 20.4 1400 ( 0.4 1400)Rx x x x 0.4 1400px . Differentiatin g w.r.t. x, we have 0.4dp
dx. Now, dpd xx dp. ( 0.4 1400) 0.16 560( 0.4)dxx
xx
When ( i) x = 400 (ii) x = 482 (iii) x = 560
1.24d 0.001d 0.84d
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138
Interpretation :
(i) When x = 400, 1d. Thus, demand function is inelastic.
(ii) When x = 482, 1dB. Thus, demand function is unitary elastic.
(iii) When x = 560, 1d. Thus, demand function is elastic.
9.4 MAXIMA AND MINIMA FOR A FUNCTION OF
SINGLE VARIABLE
Before formally defining maxima and minima for a function, let us
observe the graph of a function f(x) given below:
In the above graph of a function y = f (x)we observe that at the points A, B,
C and D the function attains extreme values. At A and B, the function is
reaching a peak while at B and C the function reaches a bottom. A andB
are said to be the points where f(x) has a maxima , while C and D are the
points where f(x) has a minima .
From the graph we make the following observations:
The function f(x) isincreasing (has an upward slope) till the point A
and starts decreasing after it reaches A.
The function f(x)is decreasing (has a downward slope) till the point C
and starts increasing after i t reaches C.
The same is true about the points B and D.
The values a, b, c, and d, at which f(x) takes extreme values are called as
extreme points or stationary points. These points are called stationary as
the derivative of the function, evaluated at thes e points is zero.
Graphically, if a tangent is drawn at these points then it is parallel to the
X-axis. In other words, the slope of the tangent at these points is zero.
It is important to remember here, that a function may or may not have any
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139 A constant function has no maxima or minima.
A linear function, which is increasing (positive slope) or decreasing
(negative slope) throughout the domain, also does not attain maxima
or minima.
Point of inflexion : The points where the f(x) has neither a maxima nor a
minima is called as point of inflexion .
Maxima
A function f(x) is said to attain a maxima or a maximum value at point x =
a if it stops to increase and begins to decrease atxa.
Minima
A function f(x) is said to attain a minima or a minimum value at a point x
= a if it stops to decrease and begins to increase at x = a .
There are different methods to find the extrema of a function:
First Derivative test for maxima and minima
Let us write down the steps involved in finding the extreme values using
the first derivative test:
Step I : Find the first derivative ()fx of f(x).
Step II: Equate the first derivative to zero. i.e. put ()fx= 0 and solve
to get the station ary points, say 12,xx x etc.
Step III: Find the sign of ()fxat a point just less than and just greater
than each of the stationary points.
Step IV : If the sign of ()fxchanges from positive to negative then f(x)
has local maxima at that point. If the sign of ()fxchanges from negative
to positive then f(x) has a local minima at that point. If there is no change
in sign of ()fxthen such a point is calle d as point of inflexion.
Solved Examples
Find all the points of maxima and minima of the function using first
derivative test.
1. 32( ) 2 4 12fx x x x
Ans: Given 32( ) 2 4 12fx x x x
Differentiating w.r.t. x,
22() 3 4 4fx x x munotes.in
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140 Equatin g() 0fx, we get
234 4 0xx
i.e.236 2 4 0xx x ( 2)(3 2) 0xx 2 and 2/ 3xx
Thus, 2/3xand x = 2 are two stationary points.
Consider the point 2/3x
We evaluate ()fx at x = – 1 and x = 0 (12 / 3 0 Q )
2( 1) 3( 1) 4( 1) 4 3 4 4 3f (1 ) 0f
and2(0) 3(0) 4(0) 4 4f (0) 0f
Since()fxchanges sign from positive to negative about the point2/3x,
f(x) has a local maxima at 2/3x.
Now consider the point x = 2
We evaluate ()fx at x = 1 and x = 3 (123Q )
2(1) 3(1) 4(1) 4 3 4 4 5f (1) 0f
and2(3) 3(3) 4(3) 4 27 12 4 11f (3) 0f
Since ()fxchanges sign from negative to positive about the point x = 2,
f(x) has a local minima at x = 2.
2. 32() 2 6fx x x
Ans: Differentiating w.r.t. x,
2( ) 6 12fx x x
Equating () 0fx, we get 261 2 0xx 6 ( 2) 0xx
x = 0 and x = 2
Thus, x = 0 and x = 2 are two stationary points.
Consider the point x = 0
We evalu ate ()fxat x = – 1 and x = 1 (101 Q )
2( 1) 6( 1) 12( 1) 6 12 18f munotes.in
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141 (1 ) 0f
and2(1) 6(1) 12(1) 6 12 6f (1) 0f
Since ()fx changes sign from positive to n egative about the point x = 0,
f(x) has a local maxima at x = 0.
Now consider the point x = 2
We evaluate ()fxat x = 1 and x = 3 (123Q )
2(1) 6(1) 12(1) 6 12 6f (1) 0f
and2(3) 6(3) 12(3) 54 36 18f (3) 0f
Since ()fxchanges sign from negative to positive about the point x = 2,
f(x) has a local minima at x = 2.
9.5 SECOND DERIVATIVE TEST FOR MAXIMA AND
MINIMA
Let y = f(x) be the given function. The s teps involved in finding the
extreme points are as follows:
STEP I : Find the first derivative of f(x).i.e.dydx
STEP II : Equate the first derivative to zero i.e. put dydx = 0 and solve to
find the stationary point s say x1, x2 etc.
Step III : Find the second derivative of f(x). i.e.22dydx.
Step IV : Evaluate 22dydx at the stationary points.
Step V : (i) If at x = xi , 22dydx> 0 then the function f(x) has a minima at x
= xi and the minimum value of f(x) is f(xi) .
(ii) If at x = xi, 22dydx< 0 then the function f(x) has a maxima at x = xi
and the maximum value of f(x) is f(xi).
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142 Solved Examples
Find all the extreme values of the f ollowing functions:
1. 312 10yx x
Ans:(i) Differentiating y w.r.t. x,
23 12dyxdx
(ii) put0dy
dx: 231 2 0x
24x 2x
x = 2 and x = – 2 are the stationary points.
(iii) Differentiating dydxw.r.t x, we have,
2
26dyxdx
(iv) At x = 2 :2
26(2) 12 0dy
dx
Thus, 312 10yx x has a minima at x = 2 and the minimum value of the
function is3
min(2) 12(2) 10 8 24 10 26y .
At 2x: 2
26( 2) 12 0dy
dx
Thus, 312 10yx x has a maxima at 2x and the maximum value of
the function is 3
max( 2) 12( 2) 10 8 24 10 6y
2. 432442 4yx x x
Ans: (i) Differentiating y w.r.t. x,
324 12 8dyxx xdx
(ii) put 0dy
dx: 3241 28 0xx x
2(4 12 8) 0xx x
2(4 4 8 8) 0xx x x
[4 ( 1) 8( 1)] 0xx x x
( 1)(4 8) 0xx x
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143 (iii)Differentiating dydxw.r.t x,
2
2
212 24 8dyxx
dx
(iv) At x = 0 :2
212(0) 24(0) 8 8 0dydx
Also at x = 2: 2
2
212(2 ) 24(2) 8 48 48 8 8 0dy
dx
Thus, at x = 0 and x = 2, the function has a minima and the minimum
value is 432
min(0) 4(0) 4(0) 24 24y
At x = 1: 2
2
212(1 ) 24(1) 8 12 24 8 4 0dy
dx
Thus, at x = 1 the function has a maxima and the maximum value is
432
max(1) 4(1) 4(1) 24 1 24 25y .
3. Find two positive numbers such that their sum is 100 and product is
maximum.
Ans: Let the two numbers be x and 100 – x.
Since we want their product to be maximum,
consider the product function2(100 ) 100yx x xx .
Differentiating w.r.t. x, 100 2dyxdx
put 0dy
dx: i.e.100 2 0x 2 100x 50x.
Now, 100 2dyxdx 2
220dy
dx
At x = 50: 2
20dy
dx there is a maxima at x = 50.
Hence the required two numbers are 50 and 100 – 50 = 50.
4. If the total revenue function and the cost function for a product are
given as 230 2Rx x and26000 6Cx x , find the profit function. At
what output is the profit maximu m? What is the maximum profit?
Ans: Given 230 2Rx x and 26000 6Cx x
The profit function is calculated as RC 2230 2 6000 6xx x x 36000 36 3xx . munotes.in
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144 Since we want to maximize π , we fol low the steps as mentioned above:
Differentiating π w.r.t. x we get,
36000 36 3ddxxdx dx 236 9dxdx
Equating ddx = 0, we have: 36 – 9x2 = 0
Thus, x2 = 4 2x .
Discarding x = –2, we have x = 2 as the stationary point.
Now, 2
2dd ddx dxdx236 9 18dxxdx
At x = 2, 2
218(2) 36 0d
dx .
Thus, the profit is maximum when 2 units of the product are produced.
The maximum profit is 3max6000 36(2) 3(2) = 6000 + 72 – 24
maxRs. 6048 .
5. If the total cost function of a firm is32/318 160Cx x x , find how
many items be produced to have the average cost, marginal cost minimum.
Also show that the average and marginal cost are equal at the minimum
average cost.
Ans: Given 32/318 160Cx x x .
Average cost:
3
2 /318 160/ 3 18 160Cx x xAC x xxx … (I)
Differentiating w.r.t x, we get
2 2/ 3 18 160 183ddAC x x xdx dx … (II)
Equating this to zero, we have,
218 0 273xx .
Differentiating (II), w.r.t . x, we get 2
2203dAC
dx atx = 27, AC is minimum and the minimum average cost is 2
27(27) / 3 18(27) 160 243 486 160 83xAC … (III)
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145 Marginal Cost
32 2/318 160 36 160dC dMC x x x x xdx dx … (IV)
Differentiating (IV) w.r.t. x, we have,
236 160 2 36ddMC x x xdx dx … (V)
Equating this to zero, we have 23 6 0 1 8xx .
Differentiating (V) w.r.t. x, we get 2
22 36 2 0ddMC xdx dx at x = 18, MC is minimum.
To show AC = MC at minimum AC
At x = 27, the MC is: 2
27(27) 36(27) 160 729 972 160 83xMC … (VI)
From (III) and (VI), we conclude that at the point of minima for AC,
AC = MC.
9.6 LET US SUM UP
In this chapter we have learn:
Marginal function of economics.
Elasticity of demand and supply.
Maxima and Minima of sing le variable function.
Maxima and Minima using second order derivative test.
9.7 UNIT END EXERCISE
1. Define elasticity of demand, elasticity of supply. Interpret at
different values graphically.
2. Write a short not on elasticity.
3. Find TR, AR, MR and for the following demand functions at the
specified point. Determine whether the demand is elastic, inelastic or
unitary.
a) 250 3pDat D = 10 b) 2100 2 0.2pD D at D = 5
c) 6 0.03pxatx = 100 d) 30 0.05px at x = 1000
e) 50 0.008pDatD = 500 f) 120 4px at x = 25
g) pxatx = /2ab h) 10 0.02pD at D = 150
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146 4.If 2300 / 4Rx , find AR, MR. What is the MR at x = 20? What is the
actual revenue from the 21st unit?
5. If3228 100 1150RD D D , find (a) AR, (b) MR, (c) MR if D = 50 and
(d) actual revenue from the 51st unit.
6. The total revenue function for a product is given as20.7 1400Rx x .
Find η at ( a) x = 500, ( b) x = 1000 and ( c) x = 1500. Interpret your answer.
7. The demand function for a commodity is 225pD and the supply
function is 24 10pS. Find dand s.
8. (a) If AR = 280, η = 2.01, find MR. (b) If MR = 104 and η = 1.2, find
MR. (c) If AR = 16, MR = 10.5 find η.
9. It is known that the demand function for a certain commodity is a linear
function. When D =1200, p = 6 and when D = 1600, p =2. Find (a)
demand function, (b) total revenue and (c) marginal revenue.
10. Find the AC, MC, MAC for the following cost function s:
a) 3210 12 4 18Cx xx b) 3
22 10 1003xCx x
c) 2200 12 145Cx x d) 158 0.7Cxx
e)320.006 0.4 20 1000Cx x x f) 323 18 42 1500Cx x x
11.If 210000.009 0.04 6 AC x xx , find (a) cost function, (b) MC. Show
that 2C x MAC x MC .
12. If 2200 5 0.05AC x x , find (a) C( x), (b) MC, (c) x, when AC = MC .
13.If 22526AC x xx , find (a) C( x), (b) MC, (c) MC when x = 5. What
is the actual cost of production for the 6th unit produced?
14. Find all the points of maxima/ minima of the following functions using
the first derivative test.
(i) 32( ) 3 12 45 75fx x x x
(ii) 32( ) 6 33 168 110fx x x x
(iii)32( ) 4 15 72 12fx x x x
(iv) 3() 4fx x x
(v) 43 2( ) 4 12 46 60 50fx x x x x
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147 15. Examine for max ima and minima the following functions:
1. 25 30 10xx 2. 3239 2 4xxx 3. 348 18xx
4. 3239 3 6xxx 5. 3258 1 5xxx 6. 3239xx
7. 3224 50xx 8. 2( 1)( 2)xx 9. 3291 5 9xx x
10. 11xx 11. ( )( )xxp x q 12. 22(3 )1x
x
13. 3258 1 6xxx 14. 322 3 36 15xx x 15. 328533xxx
16. 4224xx 17. 542552 0xxx 18. 422234 1 6xxxx
19. 43 23 10 6 50xx x 20. 54355 1 0xxx
16. Divide 50 into two positive numbers such that their product is
maximum.
17. Find two positive numbers whose prod uct is 16 and sum is minimum.
18. Divide 100 into two positive numbers such that sum of their squares is
minimum.
19. Find two positive numbers whose product is 100 and sum is minimum.
20. Divide 200 into two posit ive numbers such that their product is
maximum.
21. The total cost function of a firm for producing x unit of output is given
by 3228 300 1400Cx x x . Find the output at which the AC and MC is
minimum.
22. If the total cost function is 2C a bx cx :
(i) Find (a) AC, (b) MC. How many units are to be produced to minimize
the AC and what is the minimum AC? Show that at minimum AC, the
MC = AC.
(ii) Verify that d MC ACACdx x .
23. The profit function of a factory producing p ens is given by
20.03 900 600xx . How many pens must be produced to achieve
maximum profit? What is the profit per pen?
24. Mr. Diwakar Mishra manufactures electric switches. Based on the
product and sale of x units of switches per day, the profi t function
is280 400xx . How may switches should Mr. Mishra manufacture munotes.in
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148 and sell so as to earn maximum profit? What is the maximum profit of the
factory per day? What is the profit per switch?
25.Mr.AmitPandey manufactures transistors. His initial cost is Rs. 1000
and variable cost per transistor is 0.7 x – 20, where x is no. of transistors
produced. The total revenue function is found to be2100Rx x . Find
(i) The total cost function,
(ii) Profit function,
(iii) no. of units to be produced to gain maximum profit,
(iv) maximum profit earned Mr. Pandey and
(v) profit per transistor.
9.8 LIST OF REFERENCES
Business Mathematics by Qazi Zameeruddin (Author), Vijay K.
Khanna (Author), S.K. Bhambri (Author)
A Textbook Of Business Mathematics by HazarikaPadmalochan
Business Mathematics by S.R. Arora, Taxmann
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149 10
NUMERICAL ANALYSIS
[INTERPOLATION]
Unit Structure :
10.0 Objective
10.1 Introduction
10.2 Finite differences
10.3 Forward difference operator
10.4 Newton’s forward difference interpolation formula
10.5 Backward Difference Operator
10.6 Newton’s b ackward interpolation formula
10.7 Let us sum up
10.8 Unit end exercise
10.9 List of References
10.0 OBJECTIVE
After going through this chapter you will able to know:
Finite difference method
Forward and backward difference using finite difference.
Newton ’s interpolation formula and its used.
10.1 INTRODUCTION
There are many mathematical problems which either cannot be solved by
the exiting analytical methods or even if they can be solved, their solutions
are in complex form and may not give the desired in formation. We used
numerical methods to deal with such problems.
The branch of mathematics which deals with this aspect is called
Numerical analysis. In this chapter we are going to learn forward and
backward difference operator, relation between forward d ifference and
backward difference. We find polynomial and approximate value using
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150 10.2 FINITE DIFFERENCES
Suppose the function
takes discrete data as given below:
…… …..
The process of finding the value of
corresponding to any value of
lies
between
and
is called interpolation. Thus interpolation is the
technique of estimating the value of a function for any interm ediate value
of the independent variable. We can deal with such a situation using the
method of finite differences. We begin by deriving two important
interpolation formulae by means of forward and backward difference of a
function. We begin with some basi c concepts.
10.3 FORWARD DIFFERENCE OPERATOR
Let
be a function. Where
is the independent variable and
is
dependent variable.
In the method of finite differences is that the values of the variable
are
equally spaced. Suppose
assumes values such that the difference
between any two consecutive values is a constant, as shown below.
Where
is called the interval difference.
The corresponding values of
is given by
The difference of the value of the function
these
difference are called the first forward difference of the function
It is denoted by
(read as delta f of x zero).
Where
is called forward difference operator and
is called interval
difference.
Similarly we find second forward difference of the function
,
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151
The above table is called forward difference table. Similarly we can find
other differences.
The Shift operator E:
Let
be a function of
Let
be the
consecutive values of
. Then we define the shift operator E as
Similarly,
Relation between the operator
and
:
We have learn ,
and
Hence we can write
which gives the relation between the
operators
and
.
Note: Using the relation between the operators
and
, we can express
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152 Similarly,
Example 1: Construct a forward difference table for the following data:
0 1 2 3 4
3 2 7 24 59
Solution : Prepare the forward difference table :
0 3
-1
1 2 6
5 6
2 7 12
17 6
3 24 18
35
4 59
Example 2 : If
, construct a forward difference table
by taking
Solution: given function
by taking
i.e.
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153 Prepare the forward difference table :
0 5
1
1 6 10
11 12
2 17 22
33 12
3 50 34
67 12
4 117 46
113
5 230
Example 3: Estimate the missing value in the following table:
0 1 2 3 4
7 10 13 - 43
Solution: Since four values of
are given, we assume that
is a
polynomial of degree 3.
Therefore, the missing value is 22.
10.4 NEWTON’S FORWARD DIFFERENCE
INTERPOLATION FORMULA
Let the function
, the values of the variable
are equally spaced.
Suppose
assumes values such that the difference between any two
consecutive values is a constant, as shown below.
Where
is called the interval difference.
The corresponding values of
is given by
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154
Suppose that we want to esti mate
for
We have
Now,
This is known as Newton’s forward difference interpolation formula.
This formula can also be expressed in the form:
Example 4: Using Newton’s forward difference interpo lation formula,
estimate
2 4 6 8
4 7 11 18
Solution: First we have to prepare forward difference table:
2 4
3
4 7 1
4 2
6 11 3
7
8 18
Here,
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155 Also,
Now, by Newton’s forward difference interpolation formula,
Example 5: From the foll owing table, find the number of students who
have obtained less than 45 marks, using Newton’s forward difference
interpolation.
Marks 30-40 40-50 50-60 60-70 70-80
No. of
students 31 42 51 35 31
Solution: Here we have to prepare cumulative frequency tab le.
Below 40 Below 50 Below 60 Below 70 Below 80
(y) 31 73 124 159 190
Now construct forward difference table:
40 31
42
50 73 9
51 -25
60 124 -16 37
35 12
70 159 -4
31
80 190
Here
Also
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156 Using Newton’s forward difference interpolation formula,
Therefore, the number of students who h ave obtained marks less than 45 is
48.
Example 6: Find second degree polynomial passing through the points (0,
5), (1, 4), (2, 5)
and (3, 8) using Newton’s forward difference interpolation.
Solution: First prepare forward difference table.
0 5
-1
1 4 2
1 2 5 2
3 3 8
Here
Also,
By Newton’s forward difference interpolation formula
Therefore, the required polynomial is
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157
10.5 BACKWARD DIFFERENCE OPERATOR
Let
be a function. Where
is the independent variable and
is
dependent variable.
In the method of finite differences is that the values of the variable
are
equally spaced. Suppose
assumes values such that the difference
between any two consecutive values is a constant, as shown below.
Where
is called the interval difference.
The correspond ing values of
is given by
The difference of the value of the function
these
difference are called the first forward difference of the function
It is denoted by
(read as del f of x zero).
Where
is called backward difference operator and
is called interval
difference.
Similarly we find second forward difference of the function
,
The above table is called forward difference tabl e. Similarly we can find
other differences.
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158 The Shift operator E-1:
Let
be a function of
Let
be the
consecutive values of
. Then we define the shift operator E as
Similarly,
Relation between the operator
and
:
We have learn ,
and
Hence we can write
which gives the relation between the
operators
and
.
Example 7: Construct a backward difference table for the following data:
0 1 2 3 4
3 6 11 18 27
Solution: Prepare the backward difference table.
0 3
3
1 6 2
5
2 11 2
7
3 18 2
9
4 27
Example 8:Prepare backward difference table for
with
Solution: Given function
with
i.e.
Prepare the backward difference table.
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159
-1 -1
2
0 1 -4
-2 6
1 -1 2
0 6
2 -1 8
8 6
3 7 14
22 4 29 20
42 5 71 26
68 6 139 32
100
7 239
10.6 NEWTON’S BACKWARD INTERPOLATION
FORMULA
Let the function
, the values of the variable
are equally spaced.
Suppose
assumes values such that the difference between any two
consecutive values is a constant, as shown below.
Where
is called the interval difference.
The corresponding values of
is given by
Suppose that we want to estimate
for
We have
This is known as Newton’s backward difference interpolation formula. munotes.in
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160 This formula can also be expressed in the form:
Example 9: Using the Newton’s backward difference interpolation
formula, find
for the following data:
0 1 2 3 4
2 8 17 29 42
Solution: Prepare the backward difference table.
0 2
6 1 8 3
9 2 17 3
12
3 29 3
15 4 44
Here
Also,
Using Newton’s backward difference interpolation formula,
Example 10: By using Newton’s backward difference interpolation
formula, estimate the profit for the year 2007, from the following data.
1990 1995 2000 2005 2010
Profit
(inCrs.) 32 45 62 85 107
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161 Solution: Prepare the backward difference table.
1990 3 2
13
1995 454
17 2
2000 626
23 2
2005 85 8
31 2010 116
Here
Also,
Using Ne wton’s backward difference interpolation formula,
Example 11: Use Newton’s backward difference interpolation formula to
find
given that
Solution : Prepare the backward difference table.
0 1
5 1 6 8
13 2 19 14
27 3 46
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162 Here
Also,
Using Newton’s backward difference interpolation formula,
Therefore, the required polyno mial is
10.7 LET US SUM UP
In this chapter we have learn:
Forward difference and backward difference using finite difference.
Relation between shift operator and Delta.
Relation between shift inverse operator and Del.
Newton’s forward and ba ckward difference interpolation formula.
10.8 UNIT END EXERCISE
1. Construct the forward difference table for the following points:
a) (1,5), (2, 12), (3, 23), (4, 53)
b)
2. Construct the backward difference table for the fol lowing points:
a) (3, 12), (5, 27), (7, 45), (9, 69)
b) (1, -3), (2, 6), (3, 18), (4, 31), (5, 56), (6, 98)
3. Construct the forward difference table for the following function:
a) for
b) for
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163 4. Construct the backward differen ce table for the following function:
a) for
b) for
5. Estimate the missing value in the following tables by forward /
backward difference.
a)
1 3 5 7
1 27 - 343
b)
0 1 2 3 4
5 7 - 23 37
6. Estimate
and
from the following data using Newton’s
forward/ backward difference interpolation formula.
0 1 2 3 4
0 1 8 27 64
7. The population of a town is given below:
Year 1980 1990 2000 2010 2020
Population
(in
Thousand) 10 34 69 98 124
Estimate the population for the year 1983 using Newton’s forward
difference interpolation formula.
8. Find the number of persons getting wages less than Rs. 25 from the
following table by Newton’s forward difference interpolation.
Wages(Rs.) 0-20 20-40 40-60 60-80 80-100
No. of
Persons: 11 30 26 23 10
9. By Newton’s backward interpolation formula , estimate
for the
following data.
1 2 3 4 5
3 9 21 43 69
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164 10. By Newton’s backward interpolation formula , estimate
for the
following data.
2 4 6 8 10
7 18 32 56 88
11. From the following data estimate
by Newton’s backward
difference interpolation.
5 10 15 20
50 70 100 145
12.Use Newton’s forward difference formula to find
given that
.
13. Find
for the following table by Newton’s forward/ backward
formula:
a)
0 1 2 3 4
3 6 11 18 27
b)
0 1 2 3
1 1 7 13
c)
0 1 2 3
-1 1 1 -2
d)
1 2 3 4 5
-3 0 5 12 21
14. Multiple choice questions:
i) Let h be the finite difference, then forward difference operator is defined
as
a)
b)
c)
d)
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165 ii) Let h be the finite difference, then which of the following is true for sift
operate ?
a)
b)
c)
d)
iii) Using Newton’s forward interpolation formula for give data
then value of
is
a) 8.257
b) 8.575
c) 8.875
d) 8.375
iv) Using Newton’s backward interpolation formula for give data
then value of
is
a) 16.625
b) 15.625
c) 16.675
d) 15.525
v) Which of the following is true?
a)
b)
c)
d)
10.9 LIST OF REFERENCES
Introduction method of Numerical Analysis by S.S. Sastry
Numerical Method by Jain and Iyengar
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