Solution
= Solution
The <characteristic polynomial> is
$$
\det(A-\lambda I)=\lambda(\lambda-1).
$$
For $\lambda_1=0$ and $\lambda_2=1$, corresponding <eigenvectors> are
$$
u_1=\begin{pmatrix}2\\1\end{pmatrix},
\qquad
u_2=\begin{pmatrix}3\\1\end{pmatrix}.
$$
They are linearly independent, so the general homogeneous solution is
$$
\boxed{
z(t)=\alpha\begin{pmatrix}2\\1\end{pmatrix}
+\beta e^t\begin{pmatrix}3\\1\end{pmatrix}}.
$$