Prove that matrix A have inverse 
Let $A \in M_{n}(\mathbb{R})$ such that
$$A^2+ \alpha A+\beta I_{n}=0$$
where $\space \alpha, \beta \in \mathbb{R} \space$and$\space \beta \neq 0 $.
Prove that $A$ have inverse, and find it.

I done this way:
$A^2+\alpha A+\beta I_{n}=0$
$(A+ \alpha)A=-\beta I_{n}$
$-\frac{1}{\beta}(A+\alpha)A=I_{n} \space$ (because $\beta \neq0$)
$(-\frac{1}{\beta}A-\frac{\alpha}{\beta})A=I_{n}$
So I've assumed that $(-\frac{1}{\beta}A-\frac{\alpha}{\beta})$ is the inverse of A, because its product with A is $I_{n}$
This is correct? Thanks
 A: I am adding a just a small point for the OP that is: Your approach is a bit similar to the equation we encounter during Cayley-Hamilton Theorem. There, we have: $$A^2-tr(A)A+\det(A)I_2=0$$ and when $A$ is invertible so $\det(A)\neq0$ and then we have: $$A^{-1}=\frac{-1}{\det(A)}(A+tr(A)I_2)$$
A: What you did is correct, but if you wanna be picky about it you should prove the following
$\textbf{Lemma}:$ If $M,N$ are square matrices such that $MN=I$, then both $M$ and $N$ are non sigular and $M^{-1}=N$. First we prove that $M$ and $N$ are non singular: From $MN=I$ it follows that $\det (MN)=1$, therefore $det(M)det(N)=1$ and finally $det(M)\neq 0\neq det(N)$. So $M$ has an inverse $M^{-1}$.
To prove that $M^{-1}=N$, just multiply by $M^{-1}$ on the left of both sides of the equality $MN=I$.
A: This is not really an answer. It is a discussion of Git Gud's answer.
My immediate reaction to Git Gud's answer was: That's ridiculous! You don't need all that! And then the comments got bogged down in an argument about whether the matrices had to be square and so on.
But that is not the point. The point is that the OP's proof $-$ that $(-\frac{1}{\beta}A-\frac{\alpha}{\beta})A=I_{n} -$ can be trivially modified to show that $A(-\frac{1}{\beta}A-\frac{\alpha}{\beta})=I_{n}$. So Git Gud's answer really was needlessly complicated $-$ determinants are unnecessary here.
Perhaps my comments were off the point. If so, I hope I have made myself clearer.
