Reading about graph theory, I am presented with the case in which a square incidence submatrix of a directed graph has, in each column, only 2 non-zero entries (from the definition of the incidence matrix it is known that there will only be 2 entries for each column then these entries are $1$ and $−1$).
Here is some example incidence matrix.
$$B= \begin{pmatrix}1&0&1&0\\0&-1&-1&-1\\0&0&0&1\\-1&1&0&0 \end{pmatrix} $$
that is the incidence matrix for an oriented graph, now the submatrix of the type of interest would be:
$B_{(3,4)} = \bar B \to$
\begin{pmatrix}1&0&1\\0&-1&-1\\-1&1&0 \end{pmatrix}
We delete row $3$ and column $4$ of $B.$
Therefore, each column has a sum equal to $0$.
1. This is what I understand by column having sum equal to $0$:
For column $3$ of $\bar B=\begin{pmatrix}1\\-1\\0\end{pmatrix}=$
$$\sum_{i=1}^3 B_{(i,3)}=1+(-1)+0=\color{red}{0}$$
2. I calculate the determinant by looking for a lower triangular matrix:
$$det(\bar B)=\begin{vmatrix}1&0&1\\0&-1&-1\\-1&1&0 \end{vmatrix}=-\begin{vmatrix}1&1&0\\0&-1&-1\\-1&0&1 \end{vmatrix}=\begin{vmatrix}1&1&0\\-1&-1&0\\-1&0&1 \end{vmatrix} (row_{2}=row_{2}+row_{3})=\begin{vmatrix}0&0&0\\-1&-1&0\\-1&0&1 \end{vmatrix} (row_{1}=row_{1}+row_{2})=\color{red}{0}$$
I already know that, by property, having $1$ row or column that is a linear combination between $2$ other rows or columns of a matrix, this will have a determinant equal to $0$. Maybe in some way one column entry is a linear combination of the other ? in the form of $(-1)1+0$ for the proposed submatrix? I don't know if this property of the determinant has anything to do with it, but I think so, although my specific interest is in:
Why, in this case of a matrix (columns whose entries give 0 when added), is the determinant equal to 0? Because it's not clear to me
A square matrix $M\in M_{n\times n}(\mathbb{R})$ has determinant $0$ if and only if its row vectors (or, equivalently, its column vectors) are linearly dependent.
If we denote this vectors by $v_1,v_2,\ldots, v_n$ we have that $$1\cdot v_1+1\cdot v_2+\cdots+1\cdot v_n=0,$$
that is, a non-trivial linear combination of $v_1,\ldots,v_n$ that is equal to $0$.
For this reason, $v_1,\ldots,v_n$ are linearly dependent and $\det(M)=0$.