Well, surely you can compute it using matrix operations. But it won't be very natural. Let me instead provide you with a very similar solution (based on a similar matrix) that you'll hopefully find useful. It's not new at all (Kirchhoff, 1847) but I think it's not very well known. I first learned about it in this Wu's review paper of Potts model, p. 252. Let me reproduce the main points of the derivation.
Write $U_i$ for the potential at the site $i$ and $I_i$ for the external current flowing into the site $i$. Then continuity equation gives us
$I_i = \sum_{j \neq i} C_{ij} (U_i - U_j)$
which can be rewritten as $I_i = \sum_j A_{ij} U_j$ with
$$A_{ij} = \begin{cases} \sum_{k \neq i} C_{ik} & i = j \\ -C_{ij} & i \neq j \end{cases}$$
Now one can proceed directly to solve for $U_i$, given external current flows. But it turns out that thanks to special properties of the matrix $A$ (notice that sum entries of each row gives zero) more can be said. It turns out (read the paper for details) one can express equivalent resistance between points $k$ and $l$ as
$$R_{kl} = {{\rm det}A^{(kl)} \over {\rm det} A^{(l)}}$$ where ${A^{(kl)}}$ is $A$ with the $k^{th}$ row and $l^{th}$ column removed and ${A^{(l)}}$ is $A$ with the $l^{th}$ row and $l^{th}$ column removed.
Last remark (not related directly to your question but it would be shame not to mention it now) is that those determinants can be interpreted naturally as spanning tree polynomials in $C_{ij}$ on the given graph $G$ (with or without $(kl)$ edge) and this in turn can be computed directly from the partition function of $q \to 0$ limit of the $q$-state Potts model on the said graph $G$ with weights on the edges related to their resistances.
