JEE Main 2026 — Mathematics Algebra

The characteristic equation of a $2 \times 2$2×2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$A=[ac​bd​] is given by:

$$A^2 - \text{Tr}(A)A + \det(A)I = 0$$A2−Tr(A)A+det(A)I=0

We are given that the sum of the diagonal elements is $4$4, so $\text{Tr}(A) = a + d = 4$Tr(A)=a+d=4. Substituting this into the characteristic equation, we get:

$$A^2 - 4A + (ad - bc)I = 0$$A2−4A+(ad−bc)I=0

We are also given that the matrix satisfies:

$$A^2 - 4A + 3I = 0$$A2−4A+3I=0

Comparing the two equations, we must have:

$$ad - bc = 3$$ad−bc=3

We need to find the number of quadruples $(a, b, c, d)$(a,b,c,d) with elements from the set $\{0, 1, 2, 3, 4\}${0,1,2,3,4} such that $a + d = 4$a+d=4 and $ad - bc = 3$ad−bc=3. Let us analyze the possible values for $a$a and $d$d:

Case 1: $a = 0, d = 4$a=0,d=4

$$ad - bc = 3 \implies 0 - bc = 3 \implies bc = -3$$ad−bc=3⟹0−bc=3⟹bc=−3

Since $b, c \in \{0, 1, 2, 3, 4\}$b,c∈{0,1,2,3,4}, $bc \ge 0$bc≥0. Thus, there are no solutions.

Case 2: $a = 1, d = 3$a=1,d=3

$$ad - bc = 3 \implies 3 - bc = 3 \implies bc = 0$$ad−bc=3⟹3−bc=3⟹bc=0

For $bc = 0$bc=0, either $b = 0$b=0 or $c = 0$c=0.

If $b = 0$b=0, $c$c can take any of the $5$5 values from the set.

If $c = 0$c=0, $b$b can take any of the $5$5 values from the set.

The pair $(0, 0)$(0,0) is counted twice, so the number of pairs $(b, c)$(b,c) is $5 + 5 - 1 = 9$5+5−1=9.

Case 3: $a = 2, d = 2$a=2,d=2

$$ad - bc = 3 \implies 4 - bc = 3 \implies bc = 1$$ad−bc=3⟹4−bc=3⟹bc=1

The only possible pair from the given set is $b = 1, c = 1$b=1,c=1. This gives $1$1 solution.

Case 4: $a = 3, d = 1$a=3,d=1

$$ad - bc = 3 \implies 3 - bc = 3 \implies bc = 0$$ad−bc=3⟹3−bc=3⟹bc=0

Similar to Case 2, there are $9$9 possible pairs for $(b, c)$(b,c).

Case 5: $a = 4, d = 0$a=4,d=0

$$ad - bc = 3 \implies 0 - bc = 3 \implies bc = -3$$ad−bc=3⟹0−bc=3⟹bc=−3

Again, there are no solutions.

Adding the valid matrices from all cases, the total number of matrices is $9 + 1 + 9 = 19$9+1+9=19.

Answer: $19$19