$A^2 + A + I_n = 0 \implies$ matrix $A$ is invertible I need to proof that if
$A^2 + A + I_n = 0$ then matrix $A$ is invertible.
I can see why $A^2 + A$ is invertible, but can't find a way to proof it on $A$.
 A: $I_n=-A(A+I_n)$, hence $1= \det(I_n)= \det(-A)\det(A+I_n)=(-1)^n \det(A) \det(A+I_n).$
This gives $ \det(A) \ne 0$.
A: The inverse is $-(A+I)$, since by direct calculation
$$-(A+I)A=-A^2-A=I.$$
A: Let $Ax=0$, then $0=(A^2+A+I_n)x=I_nx=x$, hence $ker(A)=\{0\}$.
Conclusion ?
A: To show that any square matrix A is invertible, you need to show that there exists a square matrix $A^{-1}$ (called the inverse of A) such that $AA^{-1} = A^{-1}A = I_n $. That's the definition of a matrix being invertible.
Now, we can use the equation $A^2 + A + I_n = O_n$ to find a potential candidate for the inverse $A^{-1}$ such that the equation $AA^{-1} = A^{-1}A = I_n $ holds. Let us solve for $I_n$:
$A^2 + A + I_n = O_n$ ⇒ $I_n = -A^2 - A$ ⇒ $I_n = A(-A - I_n)$
Here, we have our candidate inverse matrix, which is just $A^{-1} = (-A - I_n)$. Now, you must show that it satisfies the equation $AA^{-1} = A^{-1}A = I_n$. 
Let $A^{-1} = -A - I_n$. Then $AA^{-1} = A(-A - I_n) = -A^2 - AI_n = -A^2 - A = -(A^2 + A) = I_n$ 
Similarly, $A^{-1}A = (-A - I_n)A = -A^2 - I_{n}A = -A^2 - A = I_n$.
By the definition, A is invertible. 
