My Problem is this given System of differential equations. y 1 ′ = 5 y...

My Problem is this given System of differential equations.
$$\begin{gathered} y_1^{\mathrm{\prime }}=5y_1+2y_2 \\ {y}_2^{\mathrm{\prime }}=-2y_1+y_2 \end{gathered}$$
My Approach was: again, i analyze, it must be a ordinary, linear System of equations, with both being of first-order. Than i built the corresponding Matrix as follows:
$\underset{{\overrightarrow{y}}^{\mathrm{\prime }}}{\underbrace{\left(\begin{array}{c}{y}_1^{\mathrm{\prime }}\\ y_2^{\mathrm{\prime }}\end{array}\right)}}=\underset{\mathbf{A}}{\underbrace{\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)}}\underset{\overrightarrow{y}}{\underbrace{\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right)}}$
that's why:
${\overrightarrow{y}}^{\mathrm{\prime }}=\left(\begin{array}{cc}5& 2\\ -2& 1\end{array}\right)\overrightarrow{y}$
Then I determined the eigenvalues:
they are $r_1=3$ and $r_2=3$
Knowing them, I can build the corresponding eigenvectors:
they are ${\overrightarrow{v}}_1=\left(\begin{array}{c}-1\\ +1\end{array}\right)$ and ${\overrightarrow{v}}_2=\left(\begin{array}{c}0\\ 0\end{array}\right)$
Now i plug into the equation:
$$\begin{gathered} \overrightarrow{x}=c_1{e}^{r_{1}t}\overrightarrow{v_1}+c_2{e}^{r_{2}t}\overrightarrow{v_2} \\ \overrightarrow{x}=c_1{e}^{3t}\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2{e}^{3t}\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$
this lead to my result:
$$\begin{gathered} y_1=-c_1{e}^{3t}+0c_2{e}^{3t} \\ {y}_2=c_1{e}^{3t}+0c_2{e}^{3t} \\ {y}_1=-c_1{e}^{3t} \\ {y}_2=c_1{e}^{3t} \end{gathered}$$
But I doubt it's correct. My suspect are the eigenvectors, I really don't know if they are correct. And this could have lead to a wrong solution.