JEE Main 2026 Jan 22 Shift 1, Mathematics Q22: Algebra Of Determinants

$A+A^T=O$ means $A$ is skew-symmetric: $A=\left[\begin{array}{c}0ab\\ -a0c\\ -b-c0\end{array}\right]$.

From $A\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}3\\ 3\\ 2\end{array}\right]$: $-a=3$ (twice), $c-b=2$, so $a=-3$.

From $A^2\vec{v}=A\left(\left[\begin{array}{c}3\\ 3\\ 2\end{array}\right]\right)=\left[\begin{array}{c}-3\\ 19\\ -24\end{array}\right]$: computing yields $b=3$, $c=5$ (consistency check passes).

So $A+I=\left[\begin{array}{c}1-33\\ 315\\ -3-51\end{array}\right]$, with $|A+I|=1(1+25)+3(3+15)+3(-15+3)=26+54-36=44$.

Use the determinant identities for $n=3$: $|adj(M)|=|M{|}^2$ and $|cM|=c^3|M|$.

$|2\text{adj}(A+I)|=2^3|adj(A+I)|=8\cdot 44^2$. Then $|\text{adj}(2\text{adj}(A+I))|=(8\cdot 44^2{)}^2=64\cdot 44^4=2^6\cdot (2^2\cdot 11{)}^4=2^{14}\cdot 11^4$.

So $\alpha =14,\beta =0,\gamma =4$, and $\alpha +\beta +\gamma =18$.