Answer
$e^{At}=\frac{1}{4}e^{2t}\begin{bmatrix}
e^{2t}+1 & e^{2t}-1\\
e^{2t}-1 & e^{2t}+1
\end{bmatrix}$
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
3-\lambda & 1\\
1 & 3-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
3-\lambda & 1\\
1 & 3-\lambda
\end{bmatrix}=0$
$(3-\lambda)(3-\lambda)-1=0$
$\lambda_1=4, \lambda_2=2$
The eigenvalues of $A$ are $\lambda_1 = 4$ and $\lambda_2=2$, and therefore A is nondefective.
A straightforward computation yields the following eigenvectors, which correspond respectively to $\lambda_1$ and $\lambda_2$:
$v_1=(1,1)\\
v_2=(1,-1)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
1 & 1\\
1& -1
\end{bmatrix}$ and $D = diag(4, 2)$
then $e^{At} = Se^{Dt}S^{−1}$
That is, $e^{At} -S\begin{bmatrix}
e^{4t} & 0\\
0 & e^{2t}
\end{bmatrix}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}$
Substituting:
$e^{At}=\begin{bmatrix}
1 & 1\\
1 & -1
\end{bmatrix}\begin{bmatrix}
e^{4t} & 0\\
0 & e^{2t}
\end{bmatrix}\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & -\frac{1}{2}
\end{bmatrix}=\frac{1}{4}\begin{bmatrix}
e^{4t} & e^{2t}\\
e^{4t} & -e^{2t}
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1 & -1
\end{bmatrix}=\frac{1}{4}e^{2t}\begin{bmatrix}
e^{2t}+1 & e^{2t}-1\\
e^{2t}-1 & e^{2t}+1
\end{bmatrix}$
