Check whether the following matrix is totally unimodular.
$$A=\pmatrix{0&0&0&1&0&0&0\\ 1&-1&0&0&1&0&1\\ 0&1&-1&-1&0&0&0\\ 0&0&0&1&-1&0&-1\\ 0&-1&0&0&0&0&0\\ 0&0&1&0&0&1&0\\ -1&0&0&0&0&-1&0}$$
Well, when you use the Laplace expansion for $A$, you can leave out the first row without changing the result. But, nevertheless, I still have three entries not equal to zero in the second column, which means that this submatrix is not an incidence matrix. Thus, I cannot draw a bipartite graph to show that $A$ is totally unimdoular. Am I right and how can I go on?
In view of Laplace expansion, you can remove rows 1 and 5, because each of them contains only one nonzero entry. In the remaining $5\times7$ submatrix, there are exactly one $1$ and one $-1$ in each column. Hence it is a node-arc incidence matrix, which is totally unimodular.