Let A = begin{bmatrix} -1 & 1 & -1 1 & 0 & 1 | vedclass

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(B) First,calculate the determinant of $A$: $|A| = -1(0-0) - 1(1-0) - 1(0-0) = -1$.Since $A$ is a $3 	imes 3$ matrix,the property $adj(adj(A)) = |A|^

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$4$

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(B) First,calculate the determinant of $A$: $|A| = -1(0-0) - 1(1-0) - 1(0-0) = -1$.Since $A$ is a $3 	imes 3$ matrix,the property $adj(adj(A)) = |A|^{n-2} A$ holds,where $n=3$. Thus,$adj(adj(A)) = |A|^{3-2} A = (-1)A = -A$.Next,we know $adj(A) = |A|A^{-1}$. Therefore,$adj(A)(adj(adj(A))) = (|A|A^{-1})(|A|A) = |A|^2 I = (-1)^2 I = I$.The given equation becomes $A^2 - \alpha A + \beta I = M$,where $M = \begin{bmatrix} 2 & -2 & 2 \ -2 & 0 & -1 \ 0 & 0 & -1 \end{bmatrix}$.Calculate $A^2$: $\begin{bmatrix} -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \ -1 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Substituting these into the equation: $\begin{bmatrix} 2 & -1 & 1 \ -1 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} - \alpha \begin{bmatrix} -1 & 1 & -1 \ 1 & 0 & 1 \ 0 & 0 & 1 \end{bmatrix} + \beta \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 & 2 \ -2 & 0 & -1 \ 0 & 0 & -1 \end{bmatrix}$.Comparing the elements,for the element at $(1, 2)$: $-1 - \alpha = -2 \implies \alpha = 1$.For the element at $(1, 3)$: $1 + \alpha = 2 \implies \alpha = 1$.For the element at $(3, 3)$: $1 - \alpha + \beta = -1 \implies 1 - 1 + \beta = -1 \implies \beta = -1$.Thus,$(\alpha - \beta)^2 = (1 - (-1))^2 = 2^2 = 4$.

Let $A = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ satisfy $A^2 + \alpha(adj(adj(A))) + \beta(adj(A)(adj(adj(A)))) = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$ for some $\alpha, \beta \in R$. Then $(\alpha - \beta)^2$ is equal to . . . . . . .

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