Determinant of the oriented adjacency matrix of a tree 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.
 A: I may be missing something. The determinant is zero if $v$ is a cut vertex, equivalently not a vertex of degree one, and is otherwise $\pm1$. (This all follows from standard properties of oriented incidence matrices of graphs.) The sign of the determinant with depend on the chosen ordering of vertices and edges.
A: A matrix is said to be totally unimodular if the determinant of any square submatrix of the matrix is either $0$ or $\pm 1.$ Let $G$ be a graph with incidence matrix $Q(G)$, that is, a matrix corresponding to a finite oriented directed graph of $G$. It is easily proved by induction on the order of the submatrix that $Q(G)$ is totally unimodular. The proof is taken from the book (Lemma 2.6) Bapat RB. Graphs and matrices. New York: Springer; 2010 Jul 23.
Proof: Consider the statement that any $k\times k$ submatrix of $Q(G)$ has determinant $0$ or $\pm1.$ We prove the statement by induction on $k$. Clearly the statement holds for $k=1,$ since each entry of $Q(G)$ is either $0$ or $\pm1.$ Assume the statement to be true for $k-1$ and consider a $k\times k$ submatrix $B$ of $Q(G)$. If each column of $B$ has a 1 and a $-1$, then $\det B=0.$ Also, if $B$ has a zero column, the $\det B=0.$ Now suppose $B$ has a coumn with only one nonzero entry, which must be $\pm1.$ Expand the determinant of $B$ along that column and use induction assumption to conclude that $\det B$ must be 0 or $\pm1.$ 
If $G$ is tree on $n$ vertices, then any submatrix of $Q(G)$ of order $n-1$ is nonsingular.
Proof: Consider the submatrix $X$ of $Q(G)$ formed by the rows $1,\dots, n-1.$ If we add all the rows of $X$ to the last row, then the last row of $X$ becomes the negative of the last row of $Q(G)$. Thus, if $Y$ denotes the submatrix of $Q(G)$ formed by the rows $1,\dots,n-2,n,$ then $\det X=-\det Y.$ Thu, if $\det X=0,$ then $\det Y=0.$ Continuing this way we can show that if $\det X=0$ then each $(n-1)\times (n-1)$ submatrix of $Q(G)$ must be singular. In fact, we can show that if any one of the $(n-1)\times (n-1)$ submatrices of $Q(G)$ is singular, then all them must be so. However, rank $Q(G)=n-1$ and hence at least one of the $(n-1)\times (n-1)$ submatrices of $Q(G)$ must be nonsingular.  
