**Problem:** Let $A$ be a $n \times n$ integer matrix, $\det(A) = \pm 1$. Under which conditions there exist a nested sequence of principal submatrices of size $n$ such that they all have determinant $\pm 1$ ?

We call such a sequence a $\pm 1$ Principal Minor Sequence ($\pm 1$PMS). Note that such a sequence is described by a permutation of $n$ elements. For example, consider the matrix $$ A = \left[ \begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{array} \right]. $$ The sequence $(1,2,3)$ $$ A \left[ 1 \right] = \left[ \begin{array}{c} 1 \end{array} \right], \quad A \left[ 1,2 \right] = \left[ \begin{array}{cc} 1 & 1\\ 1 & 1 \end{array} \right], \quad A \left[ 1,2,3 \right] = A. $$ is not a $\pm 1$PMS since $\det(A \left[ 1,2 \right])=0$. The sequence $(1,3,2)$ is a $\pm 1$PMS. On the other hand, the matrix $$ A = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{array} \right] $$ does not have any $\pm 1$PMS.

Does anybody know if this problem has already been studied? A related problem is to estimate the number of $\pm 1$PMS of a matrix.

**Note:** It is trivial that a matrix has no $\pm 1$PMS if all its entries in the diagonal are zero.

isa condition! $\endgroup$ – Alex Degtyarev Apr 18 '15 at 18:11