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Given a matrix $A$, size $(n-1) \times n$ over $\mathbb{R}$, whose all entries are either $1$ or $0$, my experiments show that all its solutions are formed by coordinates of $c$,$-c$ and zero (for a real constant $c$).

However, I can't find a way to prove it. Can anyone suggest a proof or construct a counter example?

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Solutions? You solve a linear system, but not a matrix... – J. M. Sep 16 '11 at 14:43
It is most probably because you're solving $Ax=0$ and the remaining $1\times n$ row is your solution transposed. Is that correct ? – user13838 Sep 16 '11 at 15:10
I meant to make your A square and full rank, sorry I guess i need more coffee :) – user13838 Sep 16 '11 at 15:17

I think you're considering the $(n-1) \times n$ matrix $A$ as the augmented matrix of a system, so you're solving $Bx = b$ where $B$ is an $(n - 1) \times (n-1)$ matrix, and all elements of $B$ and $b$ are 1's and 0's. Well, consider the case $B = \pmatrix{1 & 1 & 1\cr 0 & 1 & 0\cr 0 & 0 & 1\cr}$, $b = \pmatrix{0 \cr 1 \cr 1\cr}$, where the unique solution is $\pmatrix{-2\cr 1 \cr 1\cr}$.

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