The famous Kirchhoff's Matrix-Tree theorem counts the number of spanning trees of a connected graph, that is, the number of bases of its cycle matroid. But it appeals to vertices, that's why I do not see how to generalize it to general matroids. Also the graphs with isomorphic cycle matroids may have quite different Laplacian matrices. However, possibly there exists a known generalization?



Is there Matrix-Tree theorem for counting the bases of a connected matroid?
A broader class of matroids for which you have a Matrix Tree theorem are the [regular matroids](https://en.wikipedia.org/wiki/Regular_matroid) (those representable over every field): see, e.g., <https://arxiv.org/abs/1404.3876>.


**EDIT**: Let me actually try to give a very simple explanation of what's going on here.


Let $\mathbf{M}$ be an $n\times m$ matrix representing (i.e., its columns represent) our rank $n$ regular matroid $M$. In the case of a graph, this would be the vertex-edge incidence matrix (well, almost- we have to delete the first row of the incidence matrix to get a full rank matrix). Then the analog of the (reduced) Laplacian is given by $\mathbf{L}:=\mathbf{M}\mathbf{M}^T$. Now, the key (in fact, by a result of Tutte, equivalent) property of regular matroids is that we can choose $\mathbf{M}$ to be [totally unimodular](https://en.wikipedia.org/wiki/Unimodular_matrix), meaning every minor is $=0, \pm 1$. Let's say we've done that. Then a routine application of the [Cauchy-Binet formula](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula) shows that $\mathrm{det}(\mathbf{L})$ is equal to the number of $n$-tuples of linearly independent columns of $\mathbf{M}$, i.e., the number of bases of $M$.


Note that here we only used that the *maximal* minors of $\mathbf{M}$ are $0,\pm 1$, so maybe you can get away with slightly less than a regular matroid, I'm not sure.


