Q50 / 75 · JEE Main 2026 (22 January Shift 1) · Mathematics
Let A be a $3 \times 3$ matrix such that $\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{O}$. If $\mathrm{A}\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{l}3 \\ 3 \\ 2\end{array}\right], \mathrm{A}^{2}\left[\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{c}-3 \\ 19 \\ -24\end{array}\right]$ and $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(\mathrm{~A}+\mathrm{I})))=(2)^{\alpha} \cdot(3)^{\beta} \cdot(11)^{\gamma}, \alpha, \beta, \gamma$ are non-negative integers, then $\alpha+\beta+\gamma$ is equal to $\_\_\_\_$
Correct answer: 18
Solution
$A$ is skew-symmetric: $A = \begin{bmatrix}0 & -c & b\\c & 0 & -a\\-b & a & 0\end{bmatrix}$.
From $A\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}3\\3\\2\end{bmatrix}$: $c = 3$, $a+b = -2$.
From $A^2\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}-3\\19\\-24\end{bmatrix}$: $b = 3$, $a = -5$.
$A+I = \begin{bmatrix}1&-3&3\\3&1&5\\-3&-5&1\end{bmatrix}$, $\det(A+I) = 44$.
For $3 \times 3$ matrix: $\det(\text{adj}(M)) = (\det M)^2$.
$\det(2\,\text{adj}(A+I)) = 8 \cdot 44^2 = 8 \cdot 1936 = 15488 = 2^7 \cdot 11^2$.
$\det(\text{adj}(2\,\text{adj}(A+I))) = 15488^2 = 2^{14} \cdot 11^4$.
$\alpha = 14,\ \beta = 0,\ \gamma = 4$. $\alpha+\beta+\gamma = 18$.
From $A\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}3\\3\\2\end{bmatrix}$: $c = 3$, $a+b = -2$.
From $A^2\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}-3\\19\\-24\end{bmatrix}$: $b = 3$, $a = -5$.
$A+I = \begin{bmatrix}1&-3&3\\3&1&5\\-3&-5&1\end{bmatrix}$, $\det(A+I) = 44$.
For $3 \times 3$ matrix: $\det(\text{adj}(M)) = (\det M)^2$.
$\det(2\,\text{adj}(A+I)) = 8 \cdot 44^2 = 8 \cdot 1936 = 15488 = 2^7 \cdot 11^2$.
$\det(\text{adj}(2\,\text{adj}(A+I))) = 15488^2 = 2^{14} \cdot 11^4$.
$\alpha = 14,\ \beta = 0,\ \gamma = 4$. $\alpha+\beta+\gamma = 18$.
Hence, the correct answer is 18.