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?
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 $-A-2I$.
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.