There are two problems to deal with which must be disentangled to solve problems like these.
Both angular momentum operators are vector operators, so in some sense they "take values" in $\mathbb R^3$; you are being asked for their dot product, which should be taken within that copy of $\mathbb R^3$. You would have the same problem if you were asked to calculate the dot product $\mathbf r\cdot\mathbf p$ for a single particle without spin.
The orbital and spin angular momentum operators act on the two different factors of a tensor product of Hilbet spaces. Thus any (operator) product of a scalar orbital operator with a scalar spin operator should be interpreted as a tensor product. You would have the same problem if you were asked to calculate the product $L^2S^2$, which would need to be interpreted as $L^2\otimes S^2$.
Thus, in your case, you must read $L\cdot S$ as
$$
\mathbf{L}\cdot \mathbf{S}=\sum_{i=1}^3L_iS_i=\sum_{i=1}^3L_i\otimes S_i.
$$
To compute the matrix representation of this, you should begin with the matrix representation of each $L_i$ and $S_i$. You then compute the tensor product matrices $L_i\otimes S_i$. Finally, you add all of those matrices together to get the final result.
This is all much clearer with an example. The $z$ component, for example, is easy, since each matrix is given by
$$
L_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
\quad\text{and}\quad
S_z=\frac\hbar 2 \begin{pmatrix}1&0\\0&-1\end{pmatrix},
$$
in the bases $\{|1\rangle,|0\rangle,|-1\rangle\}$ and $\{|\tfrac12\rangle,|-\tfrac12\rangle\}$ respectively. The tensor product matrix, then, in the basis $\{|1\rangle\otimes|\tfrac12\rangle ,|0\rangle\otimes|\tfrac12\rangle ,|-1\rangle\otimes|\tfrac12\rangle , |1\rangle\otimes|-\tfrac12\rangle ,|0\rangle\otimes|-\tfrac12\rangle ,|-1\rangle\otimes|-\tfrac12\rangle \}$, is given by
$$
L_z\otimes S_z=\frac{\hbar^2} 2 \begin{pmatrix}
1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
&
0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
\\
0\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
&
-1\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}
\end{pmatrix}
=\frac{\hbar^2} 2
\begin{pmatrix}
1&0&0& 0&0&0\\0&0&0&0&0&0\\0&0&-1&0&0&0\\
0&0&0&-1&0&0\\0&0&0&0&0&0\\0&0& 0&0&0&1
\end{pmatrix}.
$$
This procedure should be repeated with both the $x$ and the $y$ components. Each of those will yield a six-by-six matrix (in this case). To get your final answer you should add all three matrices.
