Linear Algebra - Find inverse of $A$ I have this problem :
$$A =  \left(\begin{array}{ccc}
3 & -1 & 1 \\
2 & 0 & 1 \\
1 & -1 & 2 \end{array}\right) $$
1) Show that $A^3-5A^2+8A-4I=0$.
2) Using (1) To find $A^{-1}$.
I did (1) show that is correct, Usually to find inverse I use the Identity Matrix.
Anyhow for this problem, I thought of something like that :
$$A^3-5A^2+8A-4I=0 \implies  
\\A^3-5A^2+8A=4I \implies 
\\\frac{A^3-5A^2+8A}{4}=I \implies
\\\frac{A^3}{4}-\frac{5}{4}A^2+2A=I \implies
\\ A\left(\frac{A^2}{4}-\frac{5}{4}A+2I\right)=I$$
Then I think that : $$A^{-1}=\left(\frac{A^2}{4}-\frac{5}{4}A+2I\right)$$
I checked if $A^{-1}A=I$, but this is not the case.
Any ideas? Thanks!

You have 
$A^3 - 5 A^2 + 8A = 4I $
Left multiply both sides by $A^{-1}$. Use the fact that matrix multiplication is associative, and $A^{-1}A=I$,
$$
A^2 - 5A + 8I = 4A^{-1}
$$
which gives you an $A^{-1}$ of
$$
\frac{1}{4}
\begin{bmatrix}
1 & 1 & -1 \\
-3 & 5 & -1 \\
-2 & 2 & 2
\end{bmatrix}
$$
You can now check that $A A^{-1}$ is indeed $I$.
