Let A = bmatrix -1 & 1 & -1 1 & 0 & 1 0 & 0 & 1 bmatrix satisfy A^2...

The determinant of matrix $A$ is:

$|A| = -1(0 - 0) - 1(1 - 0) - 1(0 - 0) = -1$

Using the properties of the adjoint of a $3 \times 3$ matrix:

$adj(adj(A)) = |A|^{3-2}A = -A$

$adj(A)(adj(adj(A))) = adj(A)(-A) = -|A|I = -(-1)I = I$

The given equation simplifies to:

$A^2 - \alpha A + \beta I = \begin{bmatrix} 2 & -2 & 2 \\ -2 & 0 & -1 \\ 0 & 0 & -1 \end{bmatrix}$

Calculating $A^2$:

$A^2 = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Substituting $A^2$, $A$, and $I$ into the simplified 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 corresponding elements on both sides:

From the $(1,2)$ element:

$-1 - \alpha = -2 \Rightarrow \alpha = 1$

From the $(2,2)$ element:

$1 + \beta = 0 \Rightarrow \beta = -1$

Therefore, the value of $(\alpha - \beta)^2$ is:

$(1 - (-1))^2 = 2^2 = 4$

Answer: $4$