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... :( 
 A: If $A^3 = 2 I$, then you know that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A^3 v = A (A (A v) = \lambda^3 v$. But, since $A^3 = 2 I$, you have $\lambda^3 v = 2 v$ or $\lambda^3 = 2$. 
There are 3 cube roots of $2$, none of which is $1$ or $-2$. Thus, $A-I$ and $A+2I$ are invertible (they have trivial nullspaces, since 1 and -2 are not eigenvalues). 
A: We have $A^3-2I=0$ which is same as $A^3-I=I$
Recall the formula $a^3-b^3=(a-b)(***)$
A: For finding the inverse of $B_a=A-aI$ (provided it exists), write $A=B_a+aI$; then
$$
(B_a+aI)^3=B_a^3+3aB_a^2+3a^2B_a+a^3I
$$
so you have
$$
B_a^3+3aB_a^2+3a^2B_a=(2-a^3)I
$$
and multiplying by $B_a^{-1}$ gives
$$
B_a^{-1}=\frac{1}{2-a^3}(B_a^2+3aB_a+3a^2I)=
\frac{1}{2-a^3}(A^2+aA+a^2I)
$$
You see that the inverse exists provided $a^3\ne2$.
Can you do the last part?
Hint:
\begin{align}
A^2-2A+2I&=A^2-2A+A^3\\
&=A(A^2+A-2I)\\
&=A(A-I)(A+2I)
\end{align}
A: 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}
