Answer
See below
Work Step by Step
Given:
$x_1(t)=\begin{bmatrix}
\sin 3t\\
\cos 3t
\end{bmatrix}$
$x_2(t)=\begin{bmatrix}
-\cos 3t\\
\sin 3t
\end{bmatrix}$
$A=\begin{bmatrix}
0 & 3\\
-3 & 0
\end{bmatrix}$
Differentiating the given vector functions with respect to t yields,
$x_1'(t)=\begin{bmatrix}
3\cos 3t\\
-3\sin 3t
\end{bmatrix}$
$x_2'(t)=\begin{bmatrix}
3\sin 3t\\
3\cos 3t
\end{bmatrix}$
whereas
$Ax_1(t)=\begin{bmatrix}
0 & 3\\
-3 & 0
\end{bmatrix}\begin{bmatrix}
\sin 3t\\
\cos 3t
\end{bmatrix}=\begin{bmatrix}
3\cos 3t\\
-3\sin 3t
\end{bmatrix}=x_1'(t)$
$Ax_2(t)=\begin{bmatrix}
0 & 3\\
-3 & 0
\end{bmatrix}\begin{bmatrix}
-\cos 3t\\
\sin 3t
\end{bmatrix}=\begin{bmatrix}
3\sin 3t\\
3\cos 3t
\end{bmatrix}=x_2'(t)$
Hence,
$x_1'(t)=Ax_1\\
x_2'(t)=Ax_2$
so that $x_1$ and $x_2$ are indeed solutions to the given vector differential equation.
Furthermore, the Wronskian of these solutions is
$W_{[x_1,x_2]}(t)=\begin{vmatrix}
\sin 3t & -\cos 3t\\
\cos 3t & \sin 3t
\end{vmatrix}=\sin^2 3t+\cos^2 3t$
Since the Wronskian is never zero, it follows that $\{x_1, x_2\}$ is linearly independent on any interval and so is a fundamental set of solutions for the given vector differential equation. Therefore, the general solution to the system is
$x(t)=c_1\begin{bmatrix}
\sin 3t \\
\cos 3t
\end{bmatrix}+c_2\begin{vmatrix}
-\cos 3t\\
\sin 3t
\end{vmatrix}$
