Solution
= Solution
Because the <matrix-valued Laplace transform> is taken componentwise, for every $i,j$ we have
$$
\begin{aligned}
\bigl(\mathcal L\{AB\}(s))_{ij}
&=\int_0^\infty e^{-st}\sum_{k=1}^n A_{ik}B_{kj}(t)\,dt\\
&=\sum_{k=1}^n A_{ik}\int_0^\infty e^{-st}B_{kj}(t)\,dt\\
&=\bigl(A\mathcal L\{B\}(s))_{ij}.
\end{aligned}
$$
The sum is finite and $A$ is constant, so it may be moved outside the integral. Therefore
$$
\boxed{\mathcal L\{AB\}=A\mathcal L\{B\}}.
$$