Let $\alpha \beta
eq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \ \alpha & \alpha & \beta \ -\beta & \alpha & 2 \alpha\end{array} ight]$. If $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \ -\alpha & 7 & -2 \alpha \ -2 \alpha & 5 & -2 \beta\end{array} ight]$ is the matrix of…

Question

Let $\alpha \beta 
eq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \ \alpha & \alpha & \beta \ -\beta & \alpha & 2 \alpha\end{array}ight]$. If $B=\left[\begin{array}{rrr}3 \alpha & -9 & 3 \alpha \ -\alpha & 7 & -2 \alpha \ -2 \alpha & 5 & -2 \beta\end{array}ight]$ is the matrix of cofactors of the elements of $A$, then $\operatorname{det}(A B)$ is equal to :

Aicharya · Accepted Answer

Equating co-factor fo $\mathrm{A}_{21}$ $\begin{aligned} & \left(2 \alpha^2-3 \alpha\right)=\alpha \\ & \alpha=0,2 \text { (accept) } \end{aligned}$
Now, $2 \alpha^2-\alpha \beta=3 \alpha$ $\begin{aligned} & \alpha=2 \quad \beta=1 \\ & |\mathrm{AB}|=|\mathrm{A} \operatorname{cof}(\mathrm{A})|=|\mathrm{A}|^3 \\ & \mathrm{~A}=\left|\begin{array}{lll} 1 & 2 & 3 \\ 2 & 2 & 1 \\ -1 & 2 & 4 \end{array}\right|=6-2(9)+3(6)=6 \end{aligned}$

Let $\alpha \beta \neq 0$ and $A=\left[\begin{array}{rrr}\beta & \alpha & 3 \\ \alpha & \alpha & \beta \\…

Solution