Now, as to the solution using a matrix equation.
The problem is to solve $\frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dx}= \begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$.
You have determined (correctly) that the eigenvalues of the coefficient matrix are 4 and -1. That means that the matrix can be "diagonalized". That is, there exists a matrix, P, such that $P\begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}P^{-1}= \begin{pmatrix}4 & 0\\0 & -1\end{pmatrix}$.
And, of course, then. that $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}= P^{-1}\begin{pmatrix}4 & 0 \\ 0 & -1 \end{pmatrix}P$.
That "P" matrix is the matrix having the eigenvectors corresponding to eigenvalues 4 and -1 so we need to find those eigevectors.
If $\begin{pmatrix} a \\ b \end{pmatrix}$ is an eigen vector corresponding to eigenvalue 4, then we have $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= \begin{pmatrix}a+ 2b \\ 3a+ 2b\end{pmatrix}= \begin{pmatrix}4a \\ 4b\end{pmatrix}$.
So a+ 2b= 4a and 3a+ 2b= 4b. Those reduce to -3a+ 2b= 0 and 3a- 2b= 0 which both reduce to 3a= 2b. There are infinitely many solutions because there are infinitely many eigenvectors corresponding to one eigenvalue (an entire subspace). We can take, as one solution, (a, b)= (2, 3).
If $\begin{pmatrix} a \\ b \end{pmatrix}$ is an eigenvector corresponding to eigenvalue -1, then we have $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= \begin{pmatrix}a+ 2b \\ 3a+ 2b\end{pmatrix}= \begin{pmatrix}-a \\ -b\end{pmatrix}$.
So a+ 2b= -a and 3a+ 2b= -b. Those reduce to 2a+ 2b= 0 and 3a+ 3b= 0 which both reduce to a= -b. There are infinitely many solutions because there are infinitely many eigenvectors corresponding to one eigenvalue (an entire subspace). We can take, as one solution, (a, b)= (1, -1).
So $P= \begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}$ and $P^{-1}= \begin{pmatrix}\frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & -\frac{2}{5}\end{pmatrix}$
Now you can check that $P^{1}AP= \begin{pmatrix}\frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & -\frac{2}{5}\end{pmatrix}\begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}\begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}= \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}$
But then $P\begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}P^{-1}= \begin{pmatrix} 1 & 2 \\ 3 & 2\end{pmatrix}$ so we can write the differential equation $\frac{d\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}}{dx}= \begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_ 2 \end{pmatrix}$
as
$\frac{d\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}}{dx}= P\begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}P^{-1}\begin{pmatrix} y_1 \\ y_2\end{pmatrix}$
Multiply both sides by $P^{-1}$, which is a constant and can be taken inside the derivative, to get $\frac{dP^{-1}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dx}= \begin{pmatrix} 4 & 0 \\ 0 & -1 \end{pmatrix}P^{-1}\begin{pmatrix}y_1\\ y_2\end{pmatrix}$.
Let $Z= \begin{pmatrix} z_1 \\ z_2\end{pmatrix}= P^{-1}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$ and the differential equation becomes
$\frac{d\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}}{dx}= \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}$.
So we have "uncoupled" the differential equations and can write $\frac{dz_1}{dx}= 4z_1$ and $\frac{dz_2}{dx}= -z_2$ which have solutions $z_1= C_1e^{4x}$ and $z_2= C_2e^{-x}$.
All that is left is to go back to $y_1$ and $y_2$. Since $\begin{pmatrix}z_1 \\ z_2\end{pmatrix}= P^{-1}\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$ we have that $\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}= P\begin{pmatrix}z_1 \\ z_2\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}\begin{pmatrix}C_1e^{4x} \\ C_2e^{-x}\end{pmatrix}= \begin{pmatrix}2C_1e^{4x}+ C_2e^{-x} \\ 3C_1e^{4x}- C_2e^{-x}\end{pmatrix}$.
