Let alpha in(0, infty) and A=begin{bmatrix} 1 | vedclass

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(C) Given $A = \begin{bmatrix} 1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2 \end{bmatrix}$.Let $M = 2A - A^T$ and $N = A - 2A^T$. Note that $M = -N^T$,so

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$16$

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(C) Given $A = \begin{bmatrix} 1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2 \end{bmatrix}$.Let $M = 2A - A^T$ and $N = A - 2A^T$. Note that $M = -N^T$,so $\operatorname{det}(M) = \operatorname{det}(-N^T) = (-1)^3 \operatorname{det}(N) = -\operatorname{det}(N)$.The given equation is $\operatorname{det}(\operatorname{adj}(M) \cdot \operatorname{adj}(N)) = 2^8$.Using the property $\operatorname{det}(\operatorname{adj}(X)) = (\operatorname{det}(X))^{n-1}$ where $n=3$,we have $\operatorname{det}(\operatorname{adj}(M)) = (\operatorname{det}(M))^2$ and $\operatorname{det}(\operatorname{adj}(N)) = (\operatorname{det}(N))^2$.Thus,$(\operatorname{det}(M))^2 \cdot (\operatorname{det}(N))^2 = 2^8$.Since $\operatorname{det}(M) = -\operatorname{det}(N)$,we have $(-\operatorname{det}(N))^2 \cdot (\operatorname{det}(N))^2 = 2^8$,which implies $(\operatorname{det}(N))^4 = 2^8$.Therefore,$(\operatorname{det}(N))^2 = 2^4 = 16$,so $\operatorname{det}(N) = \pm 4$.Now,$N = A - 2A^T = \begin{bmatrix} 1 & 2 & \alpha \ 1 & 0 & 1 \ 0 & 1 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 2 & 0 \ 4 & 0 & 2 \ 2\alpha & 2 & 4 \end{bmatrix} = \begin{bmatrix} -1 & 0 & \alpha \ -3 & 0 & -1 \ -2\alpha & -1 & -2 \end{bmatrix}$.Calculating the determinant: $\operatorname{det}(N) = -1(0 - 1) - 0 + \alpha(3 - 0) = 1 + 3\alpha$.Setting $1 + 3\alpha = 4$ (since $\alpha > 0$),we get $3\alpha = 3$,so $\alpha = 1$.Then $A = \begin{bmatrix} 1 & 2 & 1 \ 1 & 0 & 1 \ 0 & 1 & 2 \end{bmatrix}$.$\operatorname{det}(A) = 1(0 - 1) - 2(2 - 0) + 1(1 - 0) = -1 - 4 + 1 = -4$.Finally,$(\operatorname{det}(A))^2 = (-4)^2 = 16$.

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

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