If  A is matrix of size $n\times n$  such that $A^2+A+2I=0$ , then

We have $A\left(A+I\right)=-2I$

$\Rightarrow$          $\left|A\left(A+I\right)\right|=\left|-2I\right|\Rightarrow \left|A\right|\left|A+I\right|={\left(-2\right)}^n\ne 0$ .

Thus,$\left|A\right|\ne 0.$  Also, A $\left\{-\frac{1}{2}\left(A+I\right)\right\}=I$

$\Rightarrow$ $A^{-1}=-\frac{1}{2}\left(A+I\right)$ .

Clearly $A\ne 0$  for otherwise $\left|A\right|=0$ .