GATE 2025 BT Q.63: solution and answer

GATE 2025 · BTQ.63 · 2 marks · NAT

Let $A = 103 0 k 0 10 - 1$ . If the eigenvalues of $A$ are $- 2$ , $1$ , and $2$ , then the value of $k$ is ___________. (Answer in integer)

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

k = 1

Official answer: 1Source: Official answer key, GATE 2025 BT (IIT Roorkee)

Given $A = 103 0 k 0 10 - 1$ with eigenvalues $- 2, 1, 2$ .
For any square matrix, trace $(A)$ = sum of its eigenvalues. Here $trace (A) = 1 + k + (- 1) = k$ .
Sum of the given eigenvalues $= - 2 + 1 + 2 = 1$ , so $k = 1$ .
Cross-check using $det (A)$ = product of eigenvalues. Expanding along the second row (only the $(2, 2)$ entry is nonzero there): $det (A) = k 13 1 - 1 = k (- 1 - 3) = - 4 k$ .
Product of the given eigenvalues $= (- 2) (1) (2) = - 4$ , so $- 4 k = - 4 \Rightarrow k = 1$ , confirming the trace-based result.

KEY IDEA

For any square matrix, the trace equals the sum of its eigenvalues and the determinant equals the product of its eigenvalues — two independent checks on the same unknown.

WATCH OUT FOR

Forgetting that the trace–sum-of-eigenvalues relation holds for any square matrix (not just diagonalizable ones), or making a sign slip while expanding the determinant along the row/column containing k.

From the GATE 2025 BT paper. Every question of this paper · Download the full paper · All 8 Microscopy & spectroscopy questions, with notes

GATE 2025 BT Q.63: solution and answer

Answer and solution

More Microscopy & spectroscopy questions