We are given a problem
$$\frac{du}{dt} = Au$$
where
$$ u(t) = \left[\begin{array}{r} y(t) \\ z(t) \end{array}\right] $$
and
$$\frac{dy}{dt} = 2y - z$$
$$\frac{dz}{dt} = -y + 2z$$
hence
$$ \frac{d}{dt} \left[\begin{array}{r} y \\ z \end{array}\right]= \left[\begin{array}{rr} 2 & -1 \\ -1 & 2 \end{array}\right] \left[\begin{array}{r} y \\ z \end{array}\right] $$
The teacher uses this example for a demonstration on the use of eigenvalues. If one represents $u(0)$ as a combination of eigenvalues of $A$, then the solution can be represented as a combination of $e^t$ and $e^{3t}$ which correspond to eigenvalues $\lambda_1 =1$ and $\lambda_2=3$. Hence
$$u(t) = Ce^t x_1+ De^{3t}x_2$$
$x_1$ and $x_2$ are the eigenvectors of $A$, and the constants $C$ and $D$ are determined by two initial values $y(0)$ and $z(0)$.
I am curious on how to derive the result such as
$$z(t) = Ce^{t} - De^{3t}$$
I am not able to replicate this derivation, that is, how to combine eigenvectors. exponential solutions in a way that will result in $u(t)$. The eigenvalues of $A$ are 1 and 3, and eigenvectors are $$\begin{bmatrix}1\\1\end{bmatrix}\text{ and }\begin{bmatrix}1\\-1\end{bmatrix}.$$