Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$. 
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ...  but I couldn't find anything.
 A: You have
$$B=A^3+A^2-2A=A(A-I)(A+2I)\ .$$
$A$ is invertible, and since
$$A^3-I=I\qquad\Rightarrow\qquad(A-I)(A^2+A+I)=I$$
and 
$$A^3+8I=10I\qquad\Rightarrow\qquad(A+2I)(A^2-2A+4I)=10I$$
also $(A-I)$ and $(A+2I)$ are invertible.
Now, since $B$ is the product of invertible matrices, it is invertible.
A: One of the easiest ways to prove $B$ is invertible is the following;
$$B=A^2-2A+2I$$
Using that $A^3=2I$, we can write 
$$B=A^2-2A+2I= A^2-2A + A^3 = A(A^2+A-2I)=A(A+2I)(A-I)$$
Now let us prove $A$, $A+2I$ and $A-I$ are invertible. 


*

*Supose there is $v$ such that $Av=0$, then we have $0=A^20=A^3v=2Iv=2v$. So we have $v=0$. So $A$ is invertible. 

*Supose there is $v$ such that $(A+2I)v=0$, then we have $Av=-2v$ and we have $A^3v=-8v$. So we get $-8v=A^3v=2Iv=2v$. So we have $v=0$. So $(A+2I)$ is invertible.   

*Supose there is $v$ such that $(A-I)v=0$, then we have $Av=v$ and we have $A^3v=v$. So we get $v=A^3v=2Iv=2v$. So we have $v=0$. So $(A-I)$ is invertible.
Since B is the product of three invertible matrices, B itself is invertible. 
Remark: For item 1 above, we also simply present the inverse of $A$. From $A^3=2I$, we have $A^{-1}=\frac{1}{2}A^2$.
A: A few manipulations:
$$B=A^2-2A+2I$$
$$A^3 = 2I$$
then,
$$AB = A^3 -2A^2+2A \\= 2I -2A^2+2A \\= -2(A^2-2A+2I)-2A+6I \\= -2B -2A+6I$$
Now, the fact that $A^3 = 2I$ means $A$ is invertible, because $\det(A) \neq 0$
Thus,
$$B = -2A^{-1}B-2I+6A^{-1} $$
Take $B$ to one side:
$$B(I+2A^{-1}) = -2I+6A^{-1}$$
Multiply both side with $A$:
$$B(A+2I) = -2A+6I$$
I suppose those quantities will be similar. 
A: $B=(A-I)^2+I$. Now claim that $1\pm i$ are NOT  eigen values of $A$. Otherwise , $(1\pm i)^3$ are eigen values of $A^3$ , which is NOT possible. So , $\pm i$ are NOT eigen values of $A-I$. Then , $(\pm i)^2=-1$ is NOT an eigen value of $(A-I)^2$  .
Then $0$ is NOT an eigen value of $(A-I)^2+I$  and consequently $B$ is invertible.
A: You can find the inverse directly by solving 
$$\left(\alpha A^2+\beta A +\gamma I\right)B=I.$$
If you do, you will find that $\alpha = 1/10$, $\beta = 3/10$, and $\gamma = 4/10$. Hence, $$B^{-1} = \tfrac{1}{10}A^2+\tfrac{3}{10}A+\tfrac{4}{10}I.$$
This can be easily checked by multiplying out and using the relation $A^{3}=2I$.
A: Here's something I call the "miracle method" for this type of problem. Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A - 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where in the power series expansion, the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so this becomes
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, and you'll find that, miraculously, this answer works!
A: The eigenvalues of $A$ are the cubic roots of $2$. Then
$$B=(A-(1+i)I)(A-(1-i)I)$$
is a regular matrix.
A: Hint: $B$ is invertible if and only if $0$ is not and eigenvalue of $B$.
A: $B=A^2-2A+2I$. Now multiplying the sides respectively by $A$ and $A^2$, and using the assumption $A^3=2I$ yields:
                             $$AB=BA=-2A^2+2A+2I$$
                             $$A^2B=BA^2=2A^2+2A-4I$$
Therefore
                             $$AB+A^2B=4A-2I$$
                             $$AB+2B=-2A+6I$$
By straightforward algebraic manipulation, it comes that:
                             $$A^2B+3AB+4B=BA^2+3BA+4B=10I$$
                             $$(A^2+3A+4I)B=B(A^2+3A+4I)=10I$$
Therefore
                             $$B^{-1}=(A^2+3A+4I)/10$$
