Mumbai University Solved Question Papers
Mathematics for Computer Engineering
Previous Year Question Paper with Solution
Engineering · Sem 3
May 2026 Examination
munotes.in
Mumbai
Mumbai University Solved Question Papers
Mathematics for Computer Engineering
Previous Year Question Paper with Solution
Engineering · Sem 3
May 2026 Examination
munotes.in
Mumbai
First published on munotes.in on 5 November 2026.
Published by munotes.in, Mumbai.
Model answers written and edited by the munotes.in editorial desk.
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The University does not publish an official answer key for this paper. The answers in this volume are model answers, written to show how a full-mark answer is built. They are a study aid, not an authority on what an examiner marked.
The question paper reproduced here is the paper as set by the University of Mumbai at the May 2026 examination.
The questions below are the paper as the University of Mumbai set it at the May 2026 examination, in the order it was set.
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The questions in this volume are the questions asked at the May 2026 examination, reproduced as the University of Mumbai set them, in the order it set them. Nothing has been reworded, added or left out. Only the answers are ours. See the original question paper.
Duration 2 hours · Total marks 60 · 21 questions answered
Instructions printed on the paper
How to use this volume
Solve the paper first, under exam conditions and against the clock. Then read the answers here and mark your own. Reading a solution before attempting the question feels productive and teaches very little, because recognising an answer is not the same as being able to write one.
Q.1
Attempt any Five Questions from the following · 15 Marks
Answer
For full marks, cover: the two properties used (sum = trace, product = determinant), the trace, the determinant worked by the first row, and a check through the characteristic equation.
For any square matrix A:
So nothing has to be solved. Only the trace and the determinant are needed.
A = [-2 -9 5 ; -5 -10 7 ; -9 -21 14].
Trace = -2 + (-10) + 14 = 2.
Expand |A| by the first row:
So |A| = -14 - 63 + 75 = -2.
The sum of the second-order principal minors is S₂ = (20 - 45) + (-28 + 45) + (-140 + 147) = -25 + 17 + 7 = -1. The characteristic equation is λ³ - 2λ² - λ + 2 = 0, that is (λ + 1)(λ - 1)(λ - 2) = 0. The eigenvalues are -1, 1 and 2. Their sum is 2 and their product is -2, which agrees.
Conclusion. The sum of the eigenvalues is the trace, 2, and their product is the determinant, -2. The eigenvalues themselves are -1, 1 and 2.
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