<s>[INST] Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\ 
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\ 
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then 
$$(A-\lambda I) = \begin{bmatrix} 0-\lambda & 0 & 1 \\
3 & 1-\lambda & 0 \\ 
-2&1&4-\lambda\end{bmatrix}$$
$$|A-\lambda I| = -(\lambda)^3 + 5(\lambda)^2 - 6(\lambda) +5$$
Using Cayley-Hamilton theorem
$$A^3-5^2+6A-5=0$$
How do I use this find $A^5 - 27A^3 + 65A^2$?
 [/INST] I don't see a way to find the answer directly but you can certainly simplify your calculations; since $A$ commutes with itself we may factor the polynomial however we want. Write 
$$
A^5-27A^3+65A^2=A^2(A^3-27A+65)=A^2\left[(A^3-5A^2+6A-5)+(5A^2-33A+70)\right].
$$
Now by Cayley Hamilton you only need to compute $A^2(5A^2-33A+70)$, which cuts down on the powers of $A$ that you need to calculate directly and reduces the problem to a bunch of addition and two matrix multiplications.
</s>