Let $(V,E)$ be a finite oriented directed graph, with vertices and edges ordered, and $M$ the $|V|\times |E|$ matrix with entries
$$ m_{ve} = \begin{cases} 1 &\text{if $e$ points at $v$}\\
-1 &\text{if $e$ points from $v$}\\
0 &\text{otherwise.} \end{cases} $$
If $(V,E)$ is a tree, then this matrix has one more row than being square.

> If we erase the row corresponding to a vertex $v$, the resulting square matrix is easily seen to have determinant $\pm 1$ or $0$. Is there a simple, known formula for its determinant? (Surely!)

Example: consider $1 \stackrel{1}{\to} 2 \stackrel{2}{\to} 3$, with matrix
$ \begin{pmatrix} -1&0\\ 1&-1\\ 0&1 \end{pmatrix}$. Then the three choices $v=1,2,3$ give the determinants $1,-1,1$ respectively.