ISRO CS 2025 Q.57: solution and answer

ISRO 2025 · CSQ.57 · 1 mark · MCQ

The smallest Eigen value of the matrix $A = 400364107$ is

(A)
1
(B)
3
(C)
4
(D)
6

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Answer and solution

ANSWER

The smallest eigenvalue is $4$ , option (c).

Official answer: CSource: ISRO's final answer key of 15 May 2026

The eigenvalues are the roots of $det (A - λ I) = 0$ , where $A - λ I = 4 - λ 00 3 6 - λ 4 10 7 - λ$ .
Expand the determinant along the second row, which has only one non-zero entry, $6 - λ$ (sign $+$ since it sits in position $(2, 2)$ ).
$det (A - λ I) = (6 - λ) 4 - λ 0 1 7 - λ = (6 - λ) (4 - λ) (7 - λ) .$
The roots are $λ = 4, 6, 7$ . Check: the trace is $4 + 6 + 7 = 17$ , which equals the sum of the roots, and the determinant is $4 \cdot 6 \cdot 7 = 168$ , which equals the product of the roots.
The smallest eigenvalue is $4$ , option (c). The values $1$ and $3$ are off-diagonal entries and are not eigenvalues. $6$ is an eigenvalue but not the smallest.

KEY IDEA

Expand the characteristic determinant along the row with a single non-zero entry. After that the determinant factors into three linear terms, and the roots are the eigenvalues.

WATCH OUT FOR

Reading off the off-diagonal entries $1$ or $3$ as eigenvalues, or picking the middle diagonal entry $6$ without comparing all three roots. This matrix is not triangular, but its eigenvalues still equal the diagonal entries because the determinant factors along row 2.

From the ISRO CS 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 3 ISRO Eigenvalues questions, with notes