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 = $. This makes $2A = [-2]$. The identity matrix is $$.
$1 + -2 + 1 = 0$. 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 = $. 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?