Prove that $A$ is regular matrix and find $A^{-1}$ if $A,I\in M_{n\times n}(\mathbb{R})$ and $(A+I)^3=O$

Question

Expansion of a binomial gives: $$A^3+3A^2+3A+I=0$$ We know that the matrix is regular if squared and has inverse ($detA\neq 0$). Is it possible to determine from above equation that $detA\neq 0$?

Answer 1

You are basically done.

Take $A^3+3A^2+3A+I=0$ and rearrange it to get $I = -A^3 - 3A^2 - 3A$.

We then have that

$A(-A^2-3A-3I)=I$

and so $A$ is invertible and its inverse is

$A^{-1}=-A^2-3A-3I$

Answer 2

Yes. Since: $A^3+3A^2+3A+I=0$, So the minimal polynomial of $A$, is a factor of $(x+1)^3$. Thus, the only eigenvalue of $A$ is $-1$. So, $det(A)=(-1)^n\neq 0$.

Answer 3

you can see from $$A(-A^2 - 3A - 3I) = I $$ that $$A^{-1} = -A^2 - 3A - 3I $$

since $(A+I)^3 = 0$ only eigenvalue is $-1,$ therefore the determinant of $A$ is $(-1)^n.$