<s>[INST] Proving $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$ I have a matrix A 
\begin{bmatrix}
1 & 1 & 2\\ 
 0& 2 &0 \\ 
 2&3  & 1
\end{bmatrix}
I have calculated the $A^{-1}$ 
\begin{bmatrix}
-\frac{1}{2} & -\frac{5}{4} & 2\\ 
 0& \frac{3}{4} &0 \\ 
 1&\frac{1}{4}  & -\frac{1}{2}
\end{bmatrix}
And I want to prove that $A^{-1}=-\frac{1}{6}A^{2}+\frac{2}{3}A-\frac{1}{6}I$
I tried doing all the calculations (not really productive, I know) but the result wasn't correct, or I made some calculation mistake. Is there any more productive way to prove this?
I have also found the characteristic polynomial, and it's $-\lambda ^{3}+4\lambda ^{2}-\lambda-6$
I would prefer a little tip to help me rather than the solution itself
 [/INST] Note that $\operatorname{char}_A(\lambda)=-\lambda^3+4\lambda^2-\lambda-6$. The Cayley-Hamilton Theorem then implies 
$$
-A^3+4A^2-A-6I=\mathbf 0
$$
so that
$$
A\left(-A^2+4A-I\right)=6I
$$
That is,
$$
A^{-1}=
-\frac{1}{6}A^2+\frac{2}{3}A-\frac{1}{6}I
$$
as advertised.
</s>