Given $A^{100} X = \mu X$ where $A = \begin{bmatrix} 2 & 1 & - 1 \\ 0 & -2 & -2 \\ 1 & 1 & 0 \end{bmatrix}$

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

Equation(1) becomes,

$$\phi(\lambda) = -\lambda^3 + 0\lambda^2 + \lambda + 0 = 0 \\ \phi(\lambda) = -\lambda^3 + \lambda = 0 \\ \lambda ( 1 - \lambda^2) = 0$$

$\therefore$ The eigen values of A are $\lambda = 0, - 1, 1$

Now we know that

$$A^n X = \lambda^n X \\ \therefore A^{100} X = \lambda^{100} X$$

For $\lambda$ = 0, $\mu$ = 0

For $\lambda$ = -1, $\mu$ = 1

For $\lambda$ = 1, $\mu$ = 1