Answer
$e^{At}=e^{-1}\begin{bmatrix}
\cos 3t & \sin 3t\\
-\sin 3t & \cos 3t
\end{bmatrix}$
Work Step by Step
Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-1-\lambda & 3\\
-3 & -1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
-1-\lambda & 3\\
-3 & -1-\lambda
\end{bmatrix}=0$
$(-1-\lambda)(-1-\lambda)+9=0$
$\lambda_1=-1-3i, \lambda_2=-1+3i$
The eigenvalues of $A$ are $\lambda_1 = -1-3i$ and $\lambda_2=-1+3i$, and therefore A is nondefective.
A straightforward computation yields the following eigenvectors, which correspond respectively to $\lambda_1$ and $\lambda_2$:
$v_1=(1,-i)\\
v_2=(1,i)$
It follows from Theorem 7.4.3 that if we set $S =\begin{bmatrix}
1 & 1\\
-i& i
\end{bmatrix}$ and $D = diag(-1-3i, -1+3i)$
then $e^{At} = Se^{Dt}S^{−1}$
That is, $e^{At} -S\begin{bmatrix}
e^{(-1-3i)t} & 0\\
0 & e^{(-1+3i)t}
\end{bmatrix}S^{−1}$
It is easily shown that
$S^{-1}=\begin{bmatrix}
\frac{1}{2} & \frac{i}{2}\\
\frac{1}{2} & -\frac{i}{2}
\end{bmatrix}$
Substituting:
$e^{At}=\begin{bmatrix}
1 & 1\\
-i & i
\end{bmatrix}\begin{bmatrix}
e^{(-1-3i)t} & 0\\
0 & e^{(-1+3i)t}
\end{bmatrix}\begin{bmatrix}
\frac{1}{2} & \frac{i}{2}\\
\frac{1}{2} & -\frac{i}{2}
\end{bmatrix}=\frac{1}{2}e^{-t}\begin{bmatrix}
e^{-3it}+ e^{3it}\\
-ie^{-3it} -ie^{3it}
\end{bmatrix}\begin{bmatrix}
1 & \\
1 & -i
\end{bmatrix}=\frac{1}{4}e^{-t}\begin{bmatrix}
e^{-3it}+e^{3it} & i(e^{-3it}-e^{3it})\\
-i(e^{-3it}-e^{3it}) & e^{-3it}-e^{3it}
\end{bmatrix}$
Take $\cos at=\frac{e^{at}+e^{-at}}{2}\\
\sin at=\frac{e^{at}-e^{-at}}{2} $
Hence, $e^{At}=e^{-1}\begin{bmatrix}
\cos 3t & \sin 3t\\
-\sin 3t & \cos 3t
\end{bmatrix}$
