JEE Main 2026 (Online) 8th April Evening ShiftQuestion 1
Let $A=\begin{bmatrix}\alpha&1&2\\2&3&0\\0&4&5\end{bmatrix}$ and $B=\begin{bmatrix}1&0&0\\0&-5\alpha&0\\0&4\alpha&-2\alpha\end{bmatrix}+\operatorname{adj}(A)$. If $\det(B)=66$, then $\det(\operatorname{adj}(A))$ equals :
Matrices
Solution
$\det B=66$ with $\operatorname{adj}A$ added gives (computing) $\alpha=1$ as the only solution. Then $\det A=\alpha(15)-1(10)+2(8)=15-10+16=21$ for $\alpha=1$. $\det(\operatorname{adj}A)=(\det A)^2=441$. Option (C).