Let (0,) and A=[[|gathered 1 2 1 0 1 0 1 2 gathered]]|.

Question

Let $\alpha \in (0,\infty)$ and $A=[[|\begin{gathered} 1 \quad  2 \quad  \alpha \ 1 \quad  0 \quad  1 \ 0 \quad  1 \quad  2 \end{gathered}]]|$. If $\det (adj(2A-{A}^{T})\cdot adj(A-2{A}^{T}))={2}^{8}$, then $(\det (A){)}^{2}$ is equal to:

Accepted Answer

$\begin{gathered} |adj(A-2{A}^{T})\cdot adj(2A-{A}^{T})|={2}^{8} \ P=A-2{A}^{} \ Q=2{A}^{T}-A\Rightarrow {Q}^{T}=2{A}^{T}-A=-P \ |adj(P)adj(Q)|\Rightarrow |PQ|=-{2}^{4} \ \Rightarrow |P|(-|P|)=-{2}^{4}\Rightarrow |P|=4|Q|=-4 \ |A-2{A}^{T}|=4 \ A-2{A}^{T}=[[|\begin{array}{ccc}1 \quad  2 \quad  \alpha \ 1 \quad  0 \quad  1 \ 0 \quad  1 \quad  2\end{array}]]|-2[[|\begin{array}{ccc}1 \quad  1 \quad  0 \ 2 \quad  0 \quad  1 \ \alpha  \quad  1 \quad  2\end{array}]]|=[[|\begin{array}{ccc}-1 \quad  0 \quad  \alpha \ -3 \quad  0 \quad  -1 \ -2\alpha  \quad  -1 \quad  -2\end{array}]]| \ \Rightarrow |A-2{A}^{T}|=1+3\alpha =4\Rightarrow \alpha =1\Rightarrow |A|=-4 \ \Rightarrow |A{|}^{2}=16 \end{gathered}$

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