JEE Main 2026 — Matrices Question with Solution

The characteristic equation of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is given by:

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

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

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

We are also given that the matrix satisfies:

$A^2-4A+3I=0$

Comparing the two equations, we must have:

$ad-bc=3$

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

Case 1: $a=0,d=4$
$ad-bc=3⟹0-bc=3⟹bc=-3$
Since $b,c\in \{0,1,2,3,4\}$, $bc\ge 0$. Thus, there are no solutions.

Case 2: $a=1,d=3$
$ad-bc=3⟹3-bc=3⟹bc=0$
For $bc=0$, either $b=0$ or $c=0$.
If $b=0$, $c$ can take any of the $5$ values from the set.
If $c=0$, $b$ can take any of the $5$ values from the set.
The pair $(0,0)$ is counted twice, so the number of pairs $(b,c)$ is $5+5-1=9$.

Case 3: $a=2,d=2$
$ad-bc=3⟹4-bc=3⟹bc=1$
The only possible pair from the given set is $b=1,c=1$. This gives $1$ solution.

Case 4: $a=3,d=1$
$ad-bc=3⟹3-bc=3⟹bc=0$
Similar to Case 2, there are $9$ possible pairs for $(b,c)$.

Case 5: $a=4,d=0$
$ad-bc=3⟹0-bc=3⟹bc=-3$
Again, there are no solutions.

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

Answer: $19$