# Binary equation system with more unknowns than equations

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.

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.