Show that $A$ is an invertible matrix if $\left(A+I\right)^3=0$ and find $A^{-1}$

Question

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?

Answer 1

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.

Answer 2

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 $AB=I$.

Answer 3

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

Answer 4

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.