I don't know what notation you have been using, but the following could help -just in case you haven't arrived so far.
Name the vertexes of the tetrahedron $[1],[2],[3],[4]$, for instance. Then the set of edges can be denoted by $[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]$ and the one of faces as $[1,2,3],[1,2,4],[1,3,4],[2,3,4]$.
So the corresponding groups of $0,1$ and $2$ chains are the free abelian groups on these generators:
$ \begin{eqnarray*} C_0 &=& \mathbb{Z}\langle [1],[2],[3],[4]\rangle \\ C_1 &=& \mathbb{Z}\langle [1,2],[1,3],[1,4],[2,3],[2,4],[3,4]\rangle \\ C_2 &=& \mathbb{Z}\langle [1,2,3],[1,2,4],[1,3,4],[2,3,4]\rangle \end{eqnarray*} $
and the boundary operators
$ C_2 \stackrel{\partial_2}{\longrightarrow} C_1 \stackrel{\partial_1}{\longrightarrow} C_0 $
can be computed as follows:
$ \partial_1 [1,2] = [2]-[1] \ , \qquad \partial_2 [1,2,3] = [2,3]-[1,3]+[1,2] \ , \qquad \text{etc}\dots $
So you can represent these boundary operators by matrices. For instance,
$ \partial_1 = \begin{pmatrix} -1 & -1 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix} $
and pursue your computations from this point.
For instance, if we denote by $T$ the tetrahedron, then
$ H_2 (T) = \mathrm{ker}\ \partial_2 $
and some few elementary column transformations with the $\partial_2$ matrix gives us that
$ H_2 (T) = \mathbb{Z} \langle [2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] \rangle \cong \mathbb{Z}\ . $
That is, as we already knew, of course: $H_2(T)$ is $\mathbb{Z}$. But we've got more: an explicit generator for this second homology group. Namely, the 2-cycle $[2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] $, which is, of course, the sum of the faces of the tetrahedron (with a sign).