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

The characteristic equation of matrix A is

$$|A - \lambda I| = 0 \\ (-1)^3 \lambda^3 + (-1)^2 s_{1} \lambda^2 + (-1)s_{2} \lambda + |A| = 0 .......(1)$$

Where $$s_{1} = trace(A) = (1 + (-1) + (-1)) = -1 \\ \therefore s_{1} = -1 \\ s_{2} = \begin{vmatrix} -1 & 0 \\ 0 & -1 \end{vmatrix} + \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} + \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \\ = 1 - 1 - 5 = -5 \\ \therefore s_{2} = -5 \\ |A| = 1 + 4 \\ \therefore |A| = 5$$

Equation (1) becomes ,

$$-\lambda^3 - 1 \lambda^2 + 5\lambda + 5 = 0 \\ \lambda^3 + 1 \lambda^2 - 5\lambda - 5 = 0$$

By Cayley Hamilton theorem,

$$A^3 - A^2 - 5A - 5I = 0 .......................(2) $$

$$A^2 = \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ A^2 = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A^3 = A^2 \cdot A = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & -2 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ A^3 = \begin{bmatrix} 5 & 10 & 0 \\ 10 & -5 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ \therefore A^3 + A^2 - 5A - 5I = \begin{bmatrix} 5 & 10 & 0 \\ 10 & -5 & 0 \\ 0 & 0 & -1 \end{bmatrix} + \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A^3 + A^2 - 5A - 5I = 0$$

Hence Cayley Hamilton theorem is verified.

Multiplying equation(2) by $A^{-1}$, we have

$A^2 + A - 5I - 5A^{-1} = 0 \\ 5A^{-1} = A^2 + A - 5I ............(3) \\ = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ A^{-1} = \frac{1}{5} \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -5 \end{bmatrix}$

Multiplying equation (3) by $A^{-1}$, we have

$5A^{-2} = A + I - 5A^{-1} \\ = \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - 5 \frac{1}{5} \begin{bmatrix} 1 & 2 & 0 \\ 2 & -1 & 0 \\ 0 & 0 & -5 \end{bmatrix} \\ A^{-2} = \frac{1}{5} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \end{bmatrix}$