Let α ∈(0, ∞) and A= [ … ]. If det ( adj (2 A-A^T ) · adj (A-2 A^T ) )=2^8, then ( det(A))^2 is equal to:

$$\begin{aligned} & \left|\operatorname{adj}\left(\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=28 \\ & \left|\left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)\left(2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)\right|=24 \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|\left|2 \mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right|= \pm 16 \\ & \left(\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right)^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A} \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|=\left|\mathrm{A}^{\mathrm{T}}-2 \mathrm{~A}\right| \\ & \Rightarrow\left|\mathrm{A}-2 \mathrm{~A}^{\mathrm{T}}\right|^2=16 \\ & \left|\mathrm{~A}-2 \mathrm{~A}^{\mathrm{T}}\right|= \pm 4 \\ & {\left[\begin{array}{ccc}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]-\left[\begin{array}{ccc}2 & 2 & 0 \\ 4 & 0 & 2 \\ 2 \alpha & 2 & 4\end{array}\right]} \\ & \left|\begin{array}{lll}-1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2\end{array}\right| \\ & 1+3 \alpha=4 \\ & 3 \alpha=3 \\ & \alpha=1 \\ & |\mathrm{~A}|=\left|\begin{array}{lll}1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right|=-1-3=-4 \\ & |\mathrm{~A}|^2=16\end{aligned}$$​​adj(A−2 AT)(2 A−AT)​=28​( A−2 AT)(2 A−AT)​=24​ A−2 AT​​2 A−AT​=±16( A−2 AT)T=AT−2 A​ A−2 AT​=​AT−2 A​⇒​A−2 AT​2=16​ A−2 AT​=±4​110​201​α12​​−​242α​202​024​​​−1−3−2α​00−1​α−1−2​​1+3α=43α=3α=1∣ A∣=​110​201​112​​=−1−3=−4∣ A∣2=16​