A certain matrix associated to graphs I am not very familiar with graph theory, but I need some results for my work. Thus, the question is, whether the following has already been studied and where I can find it. Let $G=(V,E)$ be an graph and let $b$ be its first Betti number. We choose $b$ directed cycles in $G$, which form a homology basis. Further, we write $\tau_j\subseteq E$ for the set of edges in the $j$-th cycle. To every edge $e$ we associate an indeterminant $T_{e}$. Now consider the $b\times b$-matrix with entries
$$a_{jk}=(-1)^{\epsilon(j,k)}\sum_{e\in\tau_j\cap\tau_k}T_e,$$
where $\epsilon(j,k)$ is $0$ if the $j$-th and the $k$-th cycle have the same direction on their intersection, and $1$ otherwise. I think, one can always choose the cycles such that $\epsilon(j,k)$ is always $0$. 
Has a matrix like this been studied in graph theory, especially its determinant and its adjugate? This should be possible by elementary combinatoric, but it would be nice to have a reference and maybe there are more known interesting facts.
 A: For a connected $G$, in the case of the cycles being chosen in the usual way --- one for each edge of $G$ outside a spanning tree $\Theta$ of $G$, this matrix can be identified with a submatrix of $A^\top A$, for $A$ the oriented incidence matrix of $G$. That is, $A$ is an $|V|\times |E|$ matrix of the map $\mathbb{R}E\to \mathbb{R}V$ given by
$$
A_{ve}=\begin{cases} 0 & v\not\in e\\ 1 &e=(vv')\\-1 &e=(v'v)\end{cases}.
$$
Choosing the ordering of $E$ so that the edges outside $\Theta$ come first, the matrix in question is the $b\times b$ submatrix of $A^\top A$ formed by its 1st $b$ rows and columns.
Note that $A^\top A-2I$ is some kind of $1,-1$-adjacency matrix of the directed line graph of $G$.
A: This is to answer a question that arose in the discussion of the 1st answer. Namely, we would like to know whether there always exists an orientation of $G$ s.t. its fundamental cycles are "coherent", in the sense that if two cycles have a common edge then its orientation equals the orientations of the cycles.
This is certainly possible if $G$ admits a spanning path $P$ (a.k.a. Hamiltonian path): take a walk on $P$ from one end to the other, and orient each edge in $P$ accordingly;
this gives a total order $<$ on $V$, and each (directed) edge $(uv)$ in $P$ satisfies $u<v$ in this order; orient each edge $(ab)$ outside $P$ in the opposite direction, i.e. $a>b$.
While not every connected graph has a spanning path, it does have a rooted normal spanning tree $P$, also known as depth-first-search (DFS) tree, also known as Trémaux tree. For $P$ (i.e. you have a partial order $<$ on $V$ induced by $P$), and for every (undirected) edge $(ab)$ of $G$, one has that either $a<b$ or $b<a$.
Now, just in the case of the spanning path, orient the edges outside $P$ in the opposite direction to the one for the edges in $P$; this induces an orientation on every fundamental cycle of $G$ (such cycles are in 1-to-1 correspondence with edges outside $P$). 
