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This problem is related to linear block codes. While trying to understand how to find a faulty bit, I stumbled upon linear equation systems with more unknowns than equations.

I have a linear equation system (binary) with 3 equations and 7 unknowns.

\begin{pmatrix} 1 & 0 & 0 & 1 & 0 & 1 & 1 & | & 1 \\ 0 & 1 & 0 & 1 & 1 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 & 1 & | & 1 \end{pmatrix}

As far as I found out, I should get 3 somehow exact values and four which are parameterized. Looking at the matrix, I would say that there is no way to "generate" more zeros.
I know that I shouldn't get infinite solutions but I should get 16 different solutions.
My problem is how to find the 16 possible solutions related to this matrix.

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Your matrix is already in row reduced echelon form. Also mod $2$, $-x=x$ and $x-y=x+y.$

So if variables going left to right across top are $x,y,z,a,b,c,d$ you have $x=a+c+d+1,\ y=a+b+c+1,\ z=b+c+d+1.$ Since there are $16$ ways to chose $1$ 0r $0$ for $a,b,c,d$ you get sixteen solutions.

Added: In a way there can be at most eight solutions because the triple $(x,y,z)$ of binary values has only eight values. I'm not familiar enough with the use of your matrix to detect a faulty bit to know more about what this means.

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