The characteristic equation is

$\begin{bmatrix} 1 - \lambda & 2 & -2 \\ -1 & 3 - \lambda & 0 \\ 0 & -2 & 1 - \lambda \end{bmatrix} = 0$

After simplification we get

$(1 - \lambda) [(3 - \lambda) (1 - \lambda)] - 2[-1 + \lambda] - 2(2) = 0 \\ (\lambda^2 - 2\lambda + 1)(3 - \lambda) + 2 - 2\lambda - 4 = 0 \\ 3\lambda^2 - 6\lambda + 3 - \lambda^3 + 2\lambda^2 - \lambda + 2 - 2\lambda - 4 = 0 \\ \therefore -\lambda^3 + 5\lambda^2 - 9\lambda + 1 = 0$

After simplification we get

$\lambda^3 + 5\lambda^2 - 9\lambda + 1 = 0$

Cayley Hamilton states that this equation is satisfied by matrix A i.e.

$A^3 + 5A^2 - 9A + 1 = 0$

Now $A^2 = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \\ = \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}$

And $A^3 = \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \\ = \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}$

$A^3 - 5A^2 + 9A + I \\ \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} - 5 \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} + 9 \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Now multiplying the above eq. by $n^{-1}$ we get

$A^2 - 5A + 9A - A^{-1} = 0 \\ \therefore A^{-1} = A^2 - 5A + 9I \\ \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} -5\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} + 9\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} $

So find $A^9$ we have $A^3$

$\therefore A^6 = A^3 \times A^3 \\ = \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} \\ = \begin{bmatrix} -313 & 124 & 452 \\ 144 & -601 & 82 \\ 82 & -22 & -231 \end{bmatrix}$

$\therefore A^9 = A^6 \times A^3 \\ \begin{bmatrix} 9953 & -24206 & -1970 \\ 5559 & -1165 & -6544 \\ -6544 & 11118 & 3409 \end{bmatrix}$

The characteristic equation is

$\begin{bmatrix} 1 - \lambda & 2 & -2 \\ -1 & 3 - \lambda & 0 \\ 0 & -2 & 1 - \lambda \end{bmatrix} = 0$

After simplification we get

$(1 - \lambda) [(3 - \lambda) (1 - \lambda)] - 2[-1 + \lambda] - 2(2) = 0 \\ (\lambda^2 - 2\lambda + 1)(3 - \lambda) + 2 - 2\lambda - 4 = 0 \\ 3\lambda^2 - 6\lambda + 3 - \lambda^3 + 2\lambda^2 - \lambda + 2 - 2\lambda - 4 = 0 \\ \therefore -\lambda^3 + 5\lambda^2 - 9\lambda + 1 = 0$

After simplification we get

$\lambda^3 + 5\lambda^2 - 9\lambda + 1 = 0$

Cayley Hamilton states that this equation is satisfied by matrix A i.e.

$A^3 + 5A^2 - 9A + 1 = 0$

Now $A^2 = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \\ = \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix}$

And $A^3 = \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \\ = \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix}$

$A^3 - 5A^2 + 9A + I \\ \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} - 5 \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} + 9 \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Now multiplying the above eq. by $n^{-1}$ we get

$A^2 - 5A + 9A - A^{-1} = 0 \\ \therefore A^{-1} = A^2 - 5A + 9I \\ \begin{bmatrix} -1 & 12 & -4 \\ -4 & 7 & 2 \\ 2 & -8 & 1 \end{bmatrix} -5\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} + 9\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} $

So find $A^9$ we have $A^3$

$\therefore A^6 = A^3 \times A^3 \\ = \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} \begin{bmatrix} -13 & 42 & -2 \\ -11 & 9 & 10 \\ 10 & -22 & -3 \end{bmatrix} \\ = \begin{bmatrix} -313 & 124 & 452 \\ 144 & -601 & 82 \\ 82 & -22 & -231 \end{bmatrix}$

$\therefore A^9 = A^6 \times A^3 \\ \begin{bmatrix} 9953 & -24206 & -1970 \\ 5559 & -1165 & -6544 \\ -6544 & 11118 & 3409 \end{bmatrix}$