Show square matrix, then matrix is invertible Question:
Show that if a square matrix $A$ satisfies the equation $A^2 + 2A + I = 0$, then $A$ must be invertible.
My work: Based on the section I read, I will treat I to be an identity matrix, which is a $1 \times 1$ matrix with a $1$ or as an square matrix with main diagonal is all ones and the rest is zero. I will also treat the $O$ as a zero matrix, which is a matrix with all zeros.
So the question wants me to show that the square matrix $A$ will make the following equation true. Okay so I pick $A$ to be $[-1]$, a $1 \times 1$ matrix with a $-1$ inside. This was out of pure luck.
This makes $A^2 = [1]$. This makes $2A = [-2]$. The identity matrix is $[1]$.
$1 + -2 + 1 = 0$. I satisfied the equation with my choice of $A$ which makes my choice of the matrix $A$ an invertible matrix.
I know matrix $A *$ the inverse of $A$ is the identity matrix.
$[-1] * inverse = [1]$. So the inverse has to be $[-1]$.
So the inverse of $A$ is $A$. 
It looks right mathematically speaking. 
Anyone can tell me how they would pick the square matrix A because I pick my matrix out of pure luck? 
 A: The inverse of $A$ is almost certainly not $A$.  You have $A^2+2A=-I$, or $A(A+2I)=-I$.  Multiplying by $-1$ you get $$A(-A-2I)=I$$
Hence the inverse of $A$ is $-A-2I$.
A: (an alternative to your approach)
if $0=A^{2}+2A+I=\left(A+I\right)\left(A-I\right)$you have that the set $\left\{+1,-1\right\}$ contains all the eigenvalues of $A$. thus $0$ is not an eigenvalue of $A$ and this is equivalent to being invertible.
Edit:
as LutzL has commented correctly, $A-I$ has to be replaced by $A+I$ and thus $\left\{-1,+1\right\}$ by $\left\{-1\right\}$. The arguments still work.
A: $$\det{A} \cdot \det{(A+2I)} =\det{[A(A+2I)]}=\det{(-I)}=\pm 1 \implies \det{A} \neq 0 \iff A \space \text{is invertible}$$ 
A: Here you have to show  $A$ is invertible. For this only you have to show $ det (A) \neq 0$. 
Now since $A$ satisfies the polynomial equation $ x^{2} + 2x + 1 = 0$ (as $ A^{2} + 2 A + I = 0$ is given), which contains non-zero constant term $1$ so $ det (A) = 1 \neq 0$. Therefore $A$ has an inverse. [Q.E.D]
** Now if you have to find $A^{-1}$, then you can proceed as follows
$A^{2} +2 A + I = 0 $ .......(1)
Since A has an inverse, so multiplying both side of equation (1) by $A^{-1}$ we have
$A^{-1} (A^{2} + 2 A + I) = A^{-1}. 0 \implies A^{-1} . A^{2} +2 A^{-1}. A + A^{-1} .I=0 \implies A + 2I + A^{-1}=0 \implies A^{-1} = -A-2I $
A: Since $A^2 + 2A + I = 0$, $(A + I)^2 = 0$.
Hence $A = -I$, for any size square matrix. 
Since the $det(-I) = 1$, the inverse exists. 
Since $I* I = I$ and $-I * -I = I$
$A = -I * -I = A$
Thus, $A*A = I$, making $A$ its own inverse.
Another solution to this problem is as follows:
$A^2 + 2A = -I$
$A(A+2)= -I$
and thus $A(-A -2I) = I$ (Since the identity is basically 1), and thus $-A -2I$ is the inverse of $A$.
From earlier conclusion $A = -I$, and thus $-(-I) - 2I = -I = A$, so it remains that $A$ is its own inverse.
