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.
Best Answer
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.