Prove that determinant of matrix equal $\pm1$ or $0$

Question

We are given square binary matrix $A_n$. Data contained by A comply the following rule: if row has any 1's then they would appear there only successively (row $(1\space 1\space0\space1 )$ is impossible). For example:
$$ \left|\left( \begin{array}_ 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)\right| = -1$$ Task is to prove that determinant of matrix equal $\pm1$ or 0.
Thank you for suggestions.

Answer 1

In short, row reduce.

In more detail: If two rows have their leftmost 1s in the same column, subtract the row with a shorter sequence of 1s from the row with the longer sequence. (If they're the same length, the rows are equal, so the determinant is zero and we're done.) This operation preserves the determinant and the condition that 1s only appear successively. Repeat until all nonzero rows have their leftmost 1s in different columns. Then permute the rows to put the matrix in echelon form; this preserves the absolute value of the determinant. Then you have a square matrix in echelon form in which all entries are 0 or 1; if it's all 1s on the diagonal then the determinant is 1, and otherwise it's 0.