How do i find sequences using linear algebra? My task is this; 
Find two sequences $\{x_n\}$ and $\{y_n\}$ such that $$x_{n+1} = x_n + 3y_n\\ y_{n+1} = 2x_n + 2y_n$$
When $x_0 = 5, y_0 = -5$.
My work so far (not sure if this is the right approach):
We are interested in finding eigenvectors and values to this system. Let $A = \begin{pmatrix}1 & 3\\2 & 2\end{pmatrix}$, $r_{n+1} = \begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}$, then $r_{n+1} = Ar_n = A^nr_0$. Finding eigenvalues/vectors to $A$ we use $det(\lambda I_2-A) = 0 \to \begin{vmatrix}\lambda - 1 & -3\\-2& \lambda - 2\end{vmatrix} = (\lambda -1)(\lambda - 2) - 6 =\\ \lambda^2 - 3\lambda -4 = (\lambda +1)(\lambda - 4) \to \lambda = \{-1, 4\}$
Solving for each eigenvalue i.e. $Av_1 = -v_1$ and $Av_2 = 4v_2$ gives us eigenvectors $v_1 =\begin{pmatrix}3&-2\end{pmatrix}, v_2 = \begin{pmatrix}1&1\end{pmatrix}$. If i did this right.
Now i'm not sure if this is the right way and certainly not to proceed so any hints and good explanations would be more than welcome.
Thanks in advance! 
 A: So now you know
$$A=\begin{pmatrix}1&3\\2&2\end{pmatrix}=\overbrace{\begin{pmatrix}3&1\\\!\!-2&1\end{pmatrix}}^{=P}\begin{pmatrix}\!\!-1&0\\0&4\end{pmatrix}\overbrace{\begin{pmatrix}\frac15&\!\!-\frac15\\\frac25&\frac35\end{pmatrix}}^{=P^{-1}}\implies$$
$$A^n=P\begin{pmatrix}(-1)^n&0\\0&4^n\end{pmatrix}P^{-1}=\frac15\begin{pmatrix}3(-1)^n+2\cdot4^n&3(-1)^{n+1}+3\cdot4^n\\2(-1)^{n+1}+2\cdot4^n&2(-1)^n+3\cdot4^n\end{pmatrix}\implies$$
The sequences are then
$$\binom{x_{n+1}}{y_{n+1}}=A^n\binom{\;5}{\!-5}=\frac15\begin{pmatrix}3(-1)^n+2\cdot4^n&3(-1)^{n+1}+3\cdot4^n\\2(-1)^{n+1}+2\cdot4^n&2(-1)^n+3\cdot4^n\end{pmatrix}\binom{\;5}{\!-5}=$$$${}$$
$$=\begin{pmatrix}6(-1)^n-4^n\\{}\\4(-1)^{n-1}-4^n\end{pmatrix}$$
A: Once you found the eigenvectors and values of $A$, writing $A$ with respect to the basis of eigenvectors gives you a diagonal matrice $A^{\prime}$ such that
$$(A^{\prime})^n=\text{diag}((-1)^n,4^n)$$
Now, you may proceed from here on without trouble (don't forget the change of basis from $A$ to $A^{\prime}$).
