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 $I$.