JEE Main 2024 (Online) 5th April Evening ShiftQuestion 1
Let $\alpha\beta\neq0$ and $A=\begin{bmatrix}\beta&\alpha&3\\\alpha&\alpha&\beta\\-\beta&\alpha&2\alpha\end{bmatrix}$. If $B=\begin{bmatrix}3\alpha&-9&3\alpha\\-\alpha&7&-2\alpha\\-2\alpha&5&-2\beta\end{bmatrix}$ is the matrix of cofactors of the elements of $A$, then $\det(AB)$ is equal to :
Determinant
Solution
Matching cofactors: $C_{12}=-(2\alpha^2+\beta^2)=-9$ and $C_{11}=2\alpha^2-\alpha\beta=3\alpha$ give $\alpha=2$, $\beta=1$, and then $|A|=6$. Since $B^T=\operatorname{adj}A$, $\det B=|A|^2$, so $\det(AB)=|A|^3=216$.