Answer
$e^{At}=\begin{bmatrix}
e^t & e^{3t}-e^t\\
0 & e^{3t}
\end{bmatrix}$
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 2\\
0 & 3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 2\\
0 & 3-\lambda
\end{bmatrix}=0$
$(1-\lambda)(3-\lambda)=0$
$\lambda_1=1, \lambda_2=3$
The eigenvalues of $A$ are $\lambda_1 = 1$ and $\lambda_2=3$, and therefore A is nondefective.
A straightforward computation yields the following eigenvectors, which correspond respectively to $\lambda_1$ and $\lambda_2$:
$v_1=(1,0)\\
v_2=(1,1)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
1 & 1\\
0& 1
\end{bmatrix}$ and $D = diag(1, 3)$
then $e^{At} = Se^{Dt}S^{−1}$
That is, $e^{At} -S\begin{bmatrix}
e^t & 0\\
0 & e^{3t}
\end{bmatrix}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}$
Substituting:
$e^{At}=\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}\begin{bmatrix}
e^t & 0\\
0 & e^{3t}
\end{bmatrix}\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
e^t & e^{3t}\\
0 & e^{3t}
\end{bmatrix}\begin{bmatrix}
1 & -1\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
e^t & e^{3t}-e^t\\
0 & e^{3t}
\end{bmatrix}$
