<s>[INST] Let A be a square matrix such that $A^3 = 2I$ Let $A$ be a square matrix such that $A^3 = 2I$
i) Prove that $A - I$ is invertible and find its inverse
ii) Prove that $A + 2I$ is invertible and find its inverse
iii) Using (i) and (ii) or otherwise, prove that $A^2 - 2A + 2I$ is invertible and find its inverse as a polynomial in $A$  
$I$ refers to identity matrix. 
Am already stucked at part i). Was going along the line of showing that $(A-I)([...]) = I$ by manipulating the equation to $A^3 - I = I$ and I got stuck... :( 
 [/INST] For parts (i) and (ii):
\begin{align}
             (A-I)^{-1} & = A^2+A+I \\
             (A+2 I)^{-1} & = \frac{1}{10}(A^2-2A+4).
\end{align}
For (iii):
$$
        A^2-2A+2I=A^2-2A+A^3=A(A+2I)(A-I)
$$
The inverse of $A$ is $\frac{1}{2}A^2$. So,
\begin{align}
     (A^2-2A+2I)^{-1}&=(A-I)^{-1}(A+2I)^{-1}A^{-1}\\
       &=\frac{1}{20}(A^2+A+I)(A^2-2A+4)A^2
\end{align}
</s>