Use the Cayley-Hamilton Theorem to find the inverse of this $3$-by-$3$ matrix $A=\begin{pmatrix}
2&0&0\\
1&3&0\\
-3&5&3
\end{pmatrix}.$
I can find the characteristic polynomial $C_a(A)= -A^3 + 8A^2-21A+18$.
How to continue from here to use the Cayley-Hamilton theorem to find the inverse matrix, $A^{-1}$?

Since we know $p(A) = 0$, you can multiply by $A^{-1}$ both sides to have:
$$ -A^2 + 8 A - 21 \, I+ 18 A^{-1} = 0,$$
can you solve for $A^{-1}$?
Spoiler:

 The answer is given by $A^{-1} = \frac{1}{18} \left( 21 I - 8 A + A^2   \right) =\left(\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{6} & \frac{1}{3} & 0 \\ \frac{7}{9} & -\frac{5}{9} & \frac{1}{3}\end{array}\right)$

