Question

Sol. [Key Concept: Use $det\ A = - 1$, $adj(adj\ A) = (det\ A)^{n - 2}A$ and $adj\ A \cdot \ A = (det\ A)I$ to reduce the expression to $A^{2} - \alpha\ A + \beta\ I = M$, then match entries.]

$${|A| = - 1 }{adj(adj(A)) = |A|^{n - 2}\text{ }A = |A|^{3 - 2} \cdot \text{ }A = |A| \cdot A }{adj(A)(adj(adj(A))) = adj(A)(|A|.A) }{= |A|(adj(A).A) }$$since $adj(A)A = |A|I$,

then $adj(A)(adj(adj(A))) = |A|^{2}I$

or $adj(A)(adj(adj(A))) = I$

solving the matrix equation

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

$$ {A^{2} - \alpha A + \beta I = M}$$

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

$$ \alpha = 1$$ and $\beta = - 1$

$$\therefore(\alpha - \beta)^{2} = 4 $$