TITLE: Can A be singular?
QUESTION [7 upvotes]: Let $A\in \mathbb{C}^{n\times n}$ satisfy
$$A^{2}+A+I=0 $$ Can A be singular?
So I have:
$$
(A-I)(A^{2}+A+I)=0\\
A^{3} = I
\\
(\det A^{3}) = \det(I)
\\
(\det A)^{3} = 1\\
\det A\neq 0
$$
So $A$ is not singular.
Is this correct the way I've done this?

REPLY [9 votes]: Yes, your solution is correct! Equivalently $A^2+A+I=0$ implies that $$A(-A-I)=I$$ and therefore $A$ is invertible and $-A-I$ is it's inverse, i.e. $A^{-1}=-A-I$.