$\det(\bar B)=0$ for $\bar B$ columned matrix with only $2$ non-null entries equal to $1$ and $-1$?

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

• Hint: what is $(1 1 1 1) B$? Commented Mar 24 at 14:01

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 $$\det(M)=0$$.