$A=\begin{pmatrix}1 &2\\3 &2\end{pmatrix}$
$y(0)=\begin{pmatrix}1\\3\end{pmatrix}$
Find the eigensystem.
$\left|\begin{pmatrix}1-\lambda &2\\3 &2-\lambda\end{pmatrix}\right| = \lambda^2-3\lambda -4$
$\lambda^2-3\lambda-4 = 0\\
(\lambda-4)(\lambda+1) = 0\\
\lambda = 4,~-1
$
$\lambda_1=4 \Rightarrow v_1 = \begin{pmatrix}2\\3\end{pmatrix}$
$\lambda_2=-1 \Rightarrow v_2 = \begin{pmatrix}1\\-1\end{pmatrix}$
$y(t) = c_1 e^{4t}\begin{pmatrix}2\\3\end{pmatrix} + c_2 e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}$
$y(0) = c_1 \begin{pmatrix}2\\3\end{pmatrix} + c_2 \begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix}$
$(c_1,c_2) = \left(\dfrac 4 5, -\dfrac 3 5\right)$
$y(t) = \dfrac 4 5 e^{4t}\begin{pmatrix}2\\3\end{pmatrix} - \dfrac 3 5 e^{-t}\begin{pmatrix}1\\-1\end{pmatrix}$
$y(t) = \dfrac 1 5 \begin{pmatrix}8 e^{4t}-3 e^{-t}\\12 e^{4t} +3 e^{-t}\end{pmatrix}$