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?
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. $A^{-1}=-A-I$.