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.