Let $\alpha \in(0, \infty)$ and $A=\left[\begin{array}{lll}1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2\end{array} ight]$. If $\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T ight) \cdot \operatorname{adj}\left(A-2 A^T ight) ight)=2^8$, then $(\operatorname{det}(A))^2$ is equal to:

Question

Let $\alpha \in(0, \infty)$ and $A=\left[\begin{array}{lll}1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2\end{array}ight]$. If $\operatorname{det}\left(\operatorname{adj}\left(2 A-A^Tight) \cdot \operatorname{adj}\left(A-2 A^Tight)ight)=2^8$, then $(\operatorname{det}(A))^2$ is equal to:

Aicharya · Accepted Answer

$\begin{aligned} & \left|\operatorname{adj}\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}ight)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}ight)ight|=28 \ & \left|\left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}ight)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}ight)ight|=24 \ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}ight|\left|2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}ight|= \pm 16 \ & \left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}ight)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A} \ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}ight|=\left|\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A}ight| \ & \Rightarrow\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}ight|^2=16 \ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}ight|= \pm 4 \ & {\left[\begin{array}{ccc}1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2\end{array}ight]-\left[\begin{array}{ccc}2 & 2 & 0 \ 4 & 0 & 2 \ 2 \alpha & 2 & 4\end{array}ight]} \ & \left|\begin{array}{lll}-1 & 0 & \alpha \ -3 & 0 & -1 \ -2 \alpha & -1 & -2\end{array}ight| \ & 1+3 \alpha=4 \ & 3 \alpha=3 \ & \alpha=1 \ & |\mathrm{~A}|=\left|\begin{array}{lll}1 & 2 & 1 \ 1 & 0 & 1 \ 0 & 1 & 2\end{array}ight|=-1-3=-4 \ & |\mathrm{~A}|^2=16\end{aligned}$

Let $\alpha \in(0, \infty)$ and $A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 &…

Solution