If $\mathbf{A}$ is a $2\times 2$ matrix that satisfies $\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}$, then $\mathbf{A}$ is invertible 
$\mathbf{A}$ is a $2\times 2$ matrix which satisfies
  $\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}$,
  where $\mathbf{I}$ is the $2\times 2$ identity matrix. Prove that $\mathbf{A}$ is invertible.
Hint: Find a matrix $\mathbf{B}$ such that $\mathbf{A}\mathbf{B}=\mathbf{I}$.

I tried substituting a variable matrix for $\mathbf{A}$ and substituting $\mathbf{I}$'s value into both the original equation and the hint, but the result was full of equations and didn't seem to help much at all. I would appreciate advice for this problem.
Thanks.
--Grace
 A: Given that $A \in$ Mat$_2(\mathbb{R})$ and $A$ satisfies the matrix equation $A^2 - 4A - 7I = O_2$, we know that $$I = {1 \over 7}A(A - 4I).$$We want to find some matrix $B$ such that $AB = I$. Well, if we look above, we can rearrange some things to give us $$\begin{align}{1 \over 7}A(A - 4I) & = A\left[{1 \over 7}(A - 4I)\right] \\ &= AB \tag{Let $1 \over 7$$(A-4I) = B$} \\&= I.\end{align}$$So $A^{-1} = B = {1 \over 7}(A-4I).$
A: If you rewrite the equation $A^2-4A-7I=0$ in a bit different way
$$
A^2-4A=7I\quad\Leftrightarrow\quad (A^2-4A)\frac17=I\quad\Leftrightarrow\quad A(A-4I)\frac17=I
$$
you may recognize the matrix $B$ here.
A: Rearrange $A^2 - 4A - 7I = 0$ to 
$$A\left(\frac{1}{7}(A - 4I)\right) = I$$
showing that $A$ is invertible.
A: To solve this type of problem, first of all, it does not need to find a matrix $B$, where $B$ satisfies $AB=I$. You can proceed as follows
Here your question is "Prove that $A$ is invertible". For this only you have to show that $det (A) \neq 0$.
Now since $A$ is a $2× 2$ matrix, satisfies the polynomial equation $x^{2} − 4x - 7 = 0$ (as $A^2-4A-7I=0$ is given) which contains non-zero constant term $(-7)$, so $det (A) =-7 \neq 0$. Therefore $A$ is invertible. [Q.E.D]
** Now if you have to find $B$, where $B$ satisfies $AB=I$, then you can proceed as follows
$ A^{2}-4 A-7 I=0 $ .......(1)
Since A has an inverse, so multiplying both side of equation (1) by $A^{-1}$ we have
$A^{-1} ( A^{2}-4 A-7 I ) = A^{-1}. 0 \implies A^{-1} . A^{2} -4 A^{-1}. A - 7 A^{-1} . I=0 \implies A - 4 I - 7 A^{-1}=0 \implies A^{-1} = \frac{1}{7} (A - 4 I) = B$.
Clearly, $AB = I$ (as $A$ is invertible, so by definition $AA^{-1}=I=A
^{-1}A$).
