A property of incidence matrix of a graph Let $G$ be an oriented graph with incidence matrix $Q$, and let $B:=[b_{ij}]$ be a $k\times k$ sub-matrix of $Q$ which is non-singular. Can there exist two distinct permutations $\sigma$ and $\sigma^\prime$ of $1,\ldots ,k$ for which both the products $b_{1\sigma (1)}\cdots b_{k\sigma (k)}$ and $ b_{1\sigma^\prime (1)}\cdots b_{k\sigma^\prime (k)}$ are non-zero ? 
 A: If a set of columns of the incidence matrix of an oriented graph is linearly independent,
then the corresponding edges form a forest. Suppose we choose $k$ columns,
and then choose $k$ rows from these to form a non-singular matrix $M$.
Claim: there is a column of $M$ with exactly one non-zero entry in it. For
otherwise each column contains a 1 and a $-1$ and so the sum of the rows of $M$ is zero.
Since $M$ is invertible, this is impossible.
Note that any two permutations with nonzero products must both use this entry of $M$
Note also that if we delete from $M$ a column with exactly one nonzero entry,
and also delete the row that contained it, the resulting $(k-1)\times(k-1)$
matrix is still non-singular. The result follows by induction on $k$.
A: MORE EDIT: The edit was wrong, the matrix I wrote down is singular. Nothing worth looking at here. 
EDIT: let $Q$ have a submatrix $$\pmatrix{1&0&0&-1\cr-1&1&0&0\cr0&-1&1&0\cr0&0&-1&1\cr}$$ The submatrix is nonsingular, and you can find a permutation that gets all the $1$s, and a different permutation that gets all the $-1$s, and the products will be equal and nonzero. 
Please ignore what follows. 
Maybe I don't understand the problem, but if $$Q=\pmatrix{0&1&1&0&1\cr0&0&1&1&0\cr0&0&0&1&1\cr0&0&0&0&1\cr0&0&0&0&0\cr}$$ then $$\pmatrix{1&0&1\cr1&1&0\cr0&1&1\cr}$$ is a non-singular $3\times3$ submatrix, and $b_{13}b_{24}b_{35}=b_{15}b_{23}b_{34}=1\ne0$$
