Let [ (0, )] and [A=[[arraylll 1 & 2 & 1 & 0 & 1 0 & 1 & 2 array]]].

Question

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

Accepted Answer

$$\begin{aligned}
& \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \
& P=A-2 A^{\top} \
& Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \
& |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \
& \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text { and }|Q|=-4 \
& \left|A-2 A^T\right|=4 \
& A-2 A^T=\left[\[\begin{array}{lll} 1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2 \end{array}\end{aligned}$$\right]-2\left[$$\begin{array}{lll} 1 & 1 & 0 \ 2 & 0 & 1 \ \alpha & 1 & 2 \end{array}$$\right]=\left[$$\begin{array}{ccc} -1 & 0 & \alpha \ -3 & 0 & -1 \ -2 \alpha & -1 & -2 \end{array}$$\right] \
\& \Rightarrow\left|A-2 A\^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \
\& \Rightarrow|A|\^2=16
\]

Solutions · 1