Show that $A$ is an invertible matrix if $\left(A+I\right)^3=0$ and find $A^{-1}$ 
If $A\in M_{n\times n}\left(\mathbb{R}\right)$ is such that $ (A+I)^3=0$, show that $A$ is an invertible matrix and find the inverse of $A$.

My idea was:
\begin{eqnarray*}
0&=&\left(A+I\right)^3\\
&=&A^3+3A^2I+3AI^2+I^3\\
&=&A^3+3A^2+3A+I,
\end{eqnarray*}
then
$$I=-A^3-3A^2-3A,$$
so
$$I=A\left(-A^2-3A-3I\right).$$
It follows that
$$A^{-1}=-A^2-3A-3I.$$
Now I found the inverse matrix, but how does this show that an inverse actually exists?
 A: Your idea is good, it works. You have shown that
$$A(-A^2-3A-3I)=I,$$
which means that $-A^2-3A-3I=A^{-1}$ by definition of the inverse. This also shows that the inverse exists, because you have constructed it already.
A: This question is quite unclear. However, if your question is: 
Given $(A+I)^3=0$, show that $A$ is invertible
Then you are on the right track. Namely, you have shown $A$ is invertible by showing there is a matrix $B$ such that $AB=I$.
A: You have found out characteristic polynomial of $A$ is which is the expansion of $(A+I)^3$ so by pre  multiplying both the sides with $A^{-1}$ you get a characteristic polynomial for its inverse  . now also you have shown that $A.x=0$ which implies the matrix is invertible. Note $x$ denote whats inside the bracket which is multiplied by A
A: All of you are right actually but for one minor overlook perhaps... the given relationship between A and I cannot be interpreted as characterstic polynomial but rather MINIMAL POLYNOMIAL because A is of order n rather than three. Keeping that in mind, proving invertibility of A is simple: all eigenvalues of A are non-zero; each being -1, and therefore, det(A)= Product of eigenvalues is non-zero as well and hence, inverse of A exists. As far as, construction of the inverse goes, may be we require more information, just like that in this case A has been assumed to be a 3 x 3 matrix.
