A is an n by n matrix such that $A^3=2I$. A is an n by n matrix such that $A^3 = 2I$.
(1) Is $A$ invertible? If it is, then what is $A^{-1}$?
(2) Is $A+I$ invertible? If it is, then what is $(I+A)^{-1}$
The original version of the question is not in English, so maybe there are some grammar mistakes, sorry in advance.
 A: $A$ quite obviously has $Nullity(A)=\emptyset$ and so is invertible. You can argue by contradiction supposing that if it had a nullspace, there exists an $x\neq 0$ such that $Ax=0 \Rightarrow A^3x=0\neq 2x=2Ix=A^3x$. One way to find the inverse, a little different (kinda worse tbh) than the one in the comments is to denote the inverse by $A^{-1}$ which we know exists from above, and go from there. We have $I=(A^{-1})^3A^3=(A^{-1})^32I=2(A^{-1})^3$. Thus we have $I=2(A^{-1})^3 \Rightarrow IA^2=2(A^{-1})^3A^2 \Rightarrow A^2=2A^{-1}\Rightarrow \frac{1}{2}A^2=A^{-1}$. We also have $A+I$ is invertible, which we can argue by showing that $0$ is not an eigenvalue. By contradiction if $0$ was an eigenvalue then we have $\exists x\neq0;(A+I)x=0x \Rightarrow Ax=-Ix \Rightarrow Ax=-x$. So $-1$ is an eigenvalue of $A$. But then we have $A^3x=(-1)^3x=-x$ but also $A^3x=2Ix=2x$ and $-x=2x$ is a contradiction since this implies that $x$ is the zero vector, which it is not by assumption. If you want to try to form a contradiction yourself, another way is to take $(A+I)x=0$, and multiply it by $A^{-1}$ and use what we know for the eigenvalue of A. If this is not enough another hint is to think of the eigenvalue of $A^{-1}$ for the vector $x$. I think this works but I only thought about it in my head sorry.
A: A matrix $B$ that satisfies $AB=BA = I$ is an inverse of $A$.
Since $A ({1 \over 2} A^2) = I$ we see that $A^{-1} = {1 \over 2} A^2$.
Suppose $x^3 = 2$, then we see that
${1 \over x+1} = {1 \over x} (1-{1 \over x} + {1 \over x^2} - {1 \over x^3} + {1 \over x^4} - {1 \over x^5} +\cdots) $, and using $x^3 = 2$ we get
$=1+(-1+{1 \over x} - {1 \over x^2}) (1-{1 \over 2} + {1 \over 2^2} - \cdots)$ and using ${x^3 \over 2} =1$ we get
$= {1 \over 3} (1-x+x^2)$.
(Check by multiplying by $1+x$.)
This suggests trying $B={1 \over 3} (I-A+A^2)$ as an inverse for $I+A$.
As above, if we multiply $(I+A)B$ we get $I$ hence
$(I+A)^{-1} = {1 \over 3} (I-A+A^2)$.
Addendum: Another approach to computing $(A+I)^{-1}$.
We want to find a polynomial $p$ such that $(x+1) p(x)$ is a non zero constant. Since $x^3=2$ we posit that a degree two polynomial will suffice. Hence we are looking for $a,b$ such that $(x+1) (1 + ax+bx^2)$ is a constant.
Expanding (and using $x^3=2$) gives
$a=-1, b=1$ which is the same as above.
