Find all matrices $A$ of order $2 \times 2$ that satisfy the equation $A^2-5A+6I = O$ 
Find all matrices $A$ of order $2 \times 2$ that satisfy the equation
$$
A^2-5A+6I = O
$$

My Attempt:
We can separate the $A$ term of the given equality:
$$
\begin{align}
A^2-5A+6I &= O\\
A^2-3A-2A+6I^2 &= O
\end{align}
$$
This implies that $A\in\{3I,2I\} = \left\{\begin{pmatrix}
3 & 0\\ 
0 & 3
\end{pmatrix},
\begin{pmatrix}
2 & 0\\ 
0 & 2
\end{pmatrix}\right\}$.
Are these the only two possible values for $A$, or are there other solutions?If there are other solutions, how can I find them?

The Cayley-Hamilton theorem states that every matrix $A$ satisfies its own characteristic polynomial; that is the polynomial for which the roots are the eigenvalues of the matrix:
$p(\lambda)=\det[A-\lambda\mathbb{I}]$.
If you view the polynomial:
$a^2-5a+6=0$,
as a characteristic polynomial with roots $a=2,3$, then any matrix with eigenvalues that are any combination of 2 or 3 will satisfy the matrix polynomial:
$A^2-5A+6\mathbb{I}=0$,
that is any matrix similar to:
$\begin{pmatrix}3 & 0\\ 0 & 3\end{pmatrix}$,$\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}$,$\begin{pmatrix}2 & 0\\ 0 & 3\end{pmatrix}$.    Note:$\begin{pmatrix}3 & 0\\ 0 & 2\end{pmatrix}$ is similar to $\begin{pmatrix}2 & 0\\ 0 & 3\end{pmatrix}$.
To see why this is true, imagine $A$ is diagonalized by some matrix $S$ to give a diagonal matrix $D$ containing the eigenvalues $D_{i,i}=e_i$, $i=1..n$, that is:
$A=SDS^{-1}$, $SS^{-1}=\mathbb{I}$.
This implies:
$A^2-5A+6\mathbb{I}=0$,
$SDS^{-1}SDS^{-1}-5SDS^{-1}+6\mathbb{I}=0$,
$S^{-1}\left(SD^2S^{-1}-5SDS^{-1}+6\mathbb{I}\right)S=0$,
$D^2-5D+6\mathbb{I}=0$,
and because $D$ is diagonal, for this to hold each diagonal entry of $D$ must satisfy this polynomial:
$D_{i,i}^2-5D_{i,i}+6=0$,
but the diagonal entries are the eigenvalues of $A$ and thus it follows that the polynomial is satisfied by $A$ iff the polynomial is satisfied by the eigenvalues of $A$.
