JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 3

Reading off A.

• $A^T e_3$ is row 3 of A: $(3,1,1)$.
• $A^T(e_1+e_3)$ is row 1 + row 3, so row 1 is $(2,1,1)$.
• $Ae_3$ is column 3: $(1,3,1)^T$.
• $A(e_1+e_3)$ gives column 1 $=(2,1,3)^T$.

So $A=\begin{bmatrix}2&1&1\\1&a&3\\3&1&1\end{bmatrix}$, with $\det A=3-a=1$, so $a=2$.

$A+I$. $\det\begin{bmatrix}3&1&1\\1&3&3\\3&1&2\end{bmatrix}=3(3)-1(-7)+1(-8)=8$.

Adjugate. For a $3\times3$ matrix $M$, $\det(\operatorname{adj}M)=(\det M)^2$. Here $\det(A^2+A)=\det A\cdot\det(A+I)=8$, so the answer is $64$.

(D)