Maths Q5 – JEE-MAIN 2026

Let

S = {A = [\begin{matrix} a & b \\ c & d \end{matrix}] : 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:

Clue 1 of 3: Cayley-Hamilton Theorem

The given matrix equation $A^{2} - 4 A + 3 I = 0$ implies that this is the characteristic equation of the matrix $A$ .

🧠 Full Solution Path

Step 1: Relate Matrix Equation to Characteristic Equation✦ Active

The given matrix equation is $A^{2} - 4 A + 3 I = 0$ . By the Cayley-Hamilton theorem, this is the characteristic equation of matrix $A$ . For a $2 \times 2$ matrix $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$ , its characteristic equation is $λ^{2} - (a + d) λ + (a d - b c) = 0$ . Comparing this with the given equation, we deduce two conditions:

a + d = 4

a d - b c = 3

We are also given that $a, b, c, d \in {0, 1, 2, 3, 4}$ .

Step 2: Determine Possible Pairs for

(a, d)

○ Expand

Using the condition $a + d = 4$ and $a, d \in {0, 1, 2, 3, 4}$ , we list all possible pairs for $(a, d)$ :

$(0, 4)$ , $(1, 3)$ , $(2, 2)$ , $(3, 1)$ , $(4, 0)$ .

💡 Teacher's Secret Hint

Ensure all combinations within the given set are considered.

Step 3: Calculate

(b, c)

pairs for each

(a, d)

case○ Expand

From $a d - b c = 3$ , we have $b c = a d - 3$ . We need to find pairs $(b, c)$ such that $b, c \in {0, 1, 2, 3, 4}$ for each $(a, d)$ case:

1. For $(a, d) = (0, 4)$ : $a d = 0$ . So $b c = 0 - 3 = - 3$ . No possible $(b, c)$ pairs since $b, c \geq 0$ . (0 matrices)

2. For $(a, d) = (1, 3)$ : $a d = 3$ . So $b c = 3 - 3 = 0$ . Possible $(b, c)$ pairs are $(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (2, 0), (3, 0), (4, 0)$ . (9 matrices)

3. For $(a, d) = (2, 2)$ : $a d = 4$ . So $b c = 4 - 3 = 1$ . Possible $(b, c)$ pair is $(1, 1)$ . (1 matrix)

4. For $(a, d) = (3, 1)$ : $a d = 3$ . So $b c = 3 - 3 = 0$ . Possible $(b, c)$ pairs are $(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (2, 0), (3, 0), (4, 0)$ . (9 matrices)

5. For $(a, d) = (4, 0)$ : $a d = 0$ . So $b c = 0 - 3 = - 3$ . No possible $(b, c)$ pairs. (0 matrices)

The total number of matrices is the sum of matrices from all valid cases: $0 + 9 + 1 + 9 + 0 = 19$ .

💡 Teacher's Secret Hint

Carefully count the pairs for $b c = 0$ to avoid overcounting or undercounting.

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Maths Question 5 – JEE-MAIN 2026

Clue 1 of 3: Cayley-Hamilton Theorem

🧠 Full Solution Path