ISRO EC 2023 Q.1: solution and answer

ISRO 2023 · ECQ.1 · 1 mark · MCQ

Given that the following matrix is singular: [[4, 2x, 0],[4, 0, 2],[12, 6, 0]]. The value of x is:

(A)
2
(B)
3
(C)
1
(D)
0

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Answer and solution

ANSWER

$x = 1$ , which is option (c).

Official answer: CSource: ISRO's revised and final answer key of 16 February 2024

A square matrix is singular exactly when its determinant is $0$ .
Expand along the first row of $4412 2 x 06 020$ : $det = 4 (0 \cdot 0 - 2 \cdot 6) - 2 x (4 \cdot 0 - 2 \cdot 12) + 0 \cdot (4 \cdot 6 - 0 \cdot 12)$ .
The first term is $4 (- 12) = - 48$ .
The second term is $- 2 x (- 24) = 48 x$ .
So $det = 48 x - 48$ .
Setting $det = 0$ gives $48 x = 48$ , hence $x = 1$ .
Check: with $x = 1$ the first row is $(4, 2, 0)$ and the third row is $(12, 6, 0) = 3 \times (4, 2, 0)$ , so the rows are linearly dependent and the matrix is singular. This is option (c).

KEY IDEA

A square matrix is singular when its determinant is zero. Expand the determinant along a row, then solve the resulting linear equation for the unknown.

WATCH OUT FOR

Sign slips in the cofactor expansion, especially the minus sign on the middle term, give $x = - 1$ . Treating the zero entry in the first row as if it contributed a term is another error.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 3 ISRO Linear algebra questions, with notes