A linear algebra problem  

I came across the above problem .Everything looks fine until the last line that says $\longrightarrow$It yields the solution   $y_1=(e^t, 0)^t$.  I do not know how it came into the picture. A little explanation will be appreciated.

Can someone point me in the right direction? Thanks in advance for your time.
 A: The $e^t$ comes from the method of solving a differential equation using the characteristic equation.  
Recall that when you have an equation $y^{'}+y = 0  $, 
you can write $r+1 = 0$    as the characteristic equation and then the solution to the differential equation is 
$y = c_1e^{rt} + c_2e^{rt}$ where $c_1$ and $c_2$ are constants.  In the case above, $r=1$ with multiplicity of $2$, therefore the general solution would be
$y = c_1e^{rt} + c_2te^{rt}$
So in the case above, you simply are multiplying the eigenvector by $e^{rt}$ where $r=1$ and that is the solution $y$.
When you have a matrix as a constant, it doesn't really change much except that you have to find the eigenvectors and normalize your eigenvector to find your solution. 
A: Since your system has only one eigenvalue $\lambda$ and it only has one eigenvector $x_1$, the general solution is given by
\begin{align}
y&=(c_1x_1+c_2(tx_1+x_2))e^{\lambda t}\\
&=c_1x_1e^{\lambda t}+c_2(tx_1+x_2)e^{\lambda t}\\
&=c_1y_1+c_2y_2
\end{align}
where $x_1$ is the eigenvector belonging to $\lambda$, $x_2$ is found by letting
$$
(\lambda I-A)x_2=x_1,
$$
and $c_1,c_2$ are just undetermined coefficients. As $x_1=[0 \ 1]'$ and $\lambda=1$, we should have 
$$
y_1=x_1e^{\lambda t}=\begin{bmatrix}1\\0
\end{bmatrix}e^{\lambda t}=\begin{bmatrix}e^{\lambda t}\\0
\end{bmatrix}=\begin{bmatrix}e^t\\0
\end{bmatrix}
$$
