$$
det(A-{\lambda}I)=0\\
\begin{vmatrix}
3-\lambda & 3 \\
-2 & -1-\lambda
\end{vmatrix}={\lambda -1}^2+2=0
$$
WRONG So
$$
\lambda_1=1 +\sqrt{2}i\\
\lambda_2=1-\sqrt{2}i
$$
$$
\text{when } \lambda=1 +\sqrt{2}i\\
\begin{vmatrix}
2-\sqrt{2}i & 3 \\
-2 & -2-\sqrt{2}i
\end{vmatrix} = \begin{vmatrix}
0 \\
0
\end{vmatrix}
$$
RREF:
$$
\begin{pmatrix}
2-\sqrt{2}i & 3 \\
0 & 0
\end{pmatrix}
\quad = \begin{pmatrix}
0 \\
0
\end{pmatrix}
\quad
$$
Let x_2=t
So we can get:
$$
{(2-\sqrt{2}i})x_1=-3x_2=-3t\\
x_1=\frac{-3t}{2-\sqrt{2}i}
$$
So
$$
t*\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad
$$
Therefore:
$$
e^{1+i\sqrt2t}\begin{pmatrix}
-1-\frac{\sqrt{2}i}{2} \\
1
\end{pmatrix}
\quad =
e^t[\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
i\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad]
$$
So, general solution:
$$
y=c_1e^t\begin{pmatrix}
-cos(\sqrt{2}t)+\frac{\sqrt{2}}{2}*sin(\sqrt{2}t) \\
cos(\sqrt{2}t)
\end{pmatrix}
\quad +
c_2e^t\begin{pmatrix}
-sin(\sqrt{2}t)-\frac{\sqrt{2}}{2}*cos(\sqrt{2}t) \\
sin(\sqrt{2}t)
\end{pmatrix}
\quad
$$
