JEE Main 2026 — 22 January, Morning Shift — Question 21

Given $A + A^T = O$, so $A$ is skew-symmetric. Let $A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$. From $A \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$

we get $-a = 3$, $-b + c = 2$. From $A^2 \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and using $A \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ we get $3a + 2b = -3$. Solving: $a = -3$, $b = 3$, $c = 5$.

So $A = \begin{bmatrix} 0 & -3 & 3 \\ 3 & 0 & 5 \\ -3 & -5 & 0 \end{bmatrix}$. Then $A+I = \begin{bmatrix} 1 & -3 & 3 \\ 3 & 1 & 5 \\ -3 & -5 & 1 \end{bmatrix}$.

Compute $|A+I| = 1(1+25) +3(3+15) +3(-15+3) = 26 + 54 -36 = 44$. Now $\det(\operatorname{adj}(2\operatorname{adj}(A+I))) = |2\operatorname{adj}(A+I)|^2 = 64|\operatorname{adj}(A+I)|^2 = 64|A+I|^4 = 64 \cdot 44^4$. Factor $44 = 4 \cdot 11 = 2^2 \cdot 11$, so $44^4 = 2^8 \cdot 11^4$.

Then $64 \cdot 44^4 = 2^6 \cdot 2^8 \cdot 11^4 = 2^{14} \cdot 11^4$. Thus $\alpha = 14$, $\beta = 0$, $\gamma = 4$.

So $\alpha + \beta + \gamma = 18$.