JEE Main 2026 Apr 4 Shift 1, Mathematics Q5: Algebra Of Matrices

JEE Main2026Apr 4, Shift 1Mathematics

Let $S = {A = [ab c d] : a, b, c, d \in {0, 1, 2, 3, 4} and A^{2} - 4 A + 3 I = 0}$ be a set of $2 \times 2$ matrices. Then the number of matrices in $S$ , for which the sum of the diagonal elements is equal to $4$ , is:

A
$20$
B
$17$
C
$21$
D
$19$

Solution

By Cayley-Hamilton, $A^{2} - (tr A) A + (det A) I = 0$ . Comparing with $A^{2} - 4 A + 3 I = 0$ gives $tr A = a + d = 4$ and $det A = a d - b c = 3$ .

Run through the cases for $(a, d)$ with $a + d = 4$ and $a, d \in {0, 1, 2, 3, 4}$ :

$∙ (a, d) = (0, 4)$ : $a d - b c = - b c = 3 \Rightarrow b c = - 3$ . Impossible since $b, c \geq 0$ .

$∙ (a, d) = (1, 3)$ : $b c = 0$ . Either $b = 0$ ( $5$ choices for $c$ ) or $c = 0$ ( $5$ choices for $b$ ), minus the double-count $b = c = 0$ : $9$ matrices.

$∙ (a, d) = (2, 2)$ : $b c = 1$ . Only $(b, c) = (1, 1)$ : $1$ matrix.

$∙ (a, d) = (3, 1)$ : same as $(1, 3)$ : $9$ matrices.

$∙ (a, d) = (4, 0)$ : $b c = - 3$ . Impossible.

Total: $9 + 1 + 9 = 19.$

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JEE Main 2026 Apr 4 Shift 1, Mathematics Q5: Algebra Of Matrices

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