Let \(\mathrm{S}=\left\{\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} \in\{0,1,2,3,4\}\right.\) and \(\left.\mathrm{A}^{2}-4 \mathrm{~A}+3 \mathrm{I}=0\right\}\) 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:

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Matrices

Let

S = {A = [a c b 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:

Answer: (D)

With trace

a + d = 4

, the Cayley-Hamilton relation reads

A^{2} - 4 A + (det A) I = 0

, so

A^{2} - 4 A + 3 I = 0

forces

det A = 3

, i.e.

a d - b c = 3

. Taking

a + d = 4

with entries in

{0, 1, 2, 3, 4}

: for

a = 0, d = 4

a = 4, d = 0

one needs

b c = - 3

, impossible; for

a = 1, d = 3

a = 3, d = 1

one needs

b c = 0

, giving

9

choices each (either

b = 0

c = 0

); for

a = 2, d = 2

one needs

b c = 1

, giving the single choice

b = c = 1

. The count is

9 + 1 + 9 = 19

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option