I have a problem in which I am trying to prove that a binary symmetric reflexive matrix is always consistent in a system mod 2 where the matrix is augmented with $$ \left[\begin{array} \\ 1 \\ \vdots \\ 1 \end{array}\right] $$ Such matrix is constructed like so: $$ \left[\begin{array}{rrrrr|r} 1 & a_{1,1} & a_{1,2} & \dots & a_{1,n} & 1 \\ a_{1,1} & 1 & a_{2,1} & \ddots & \vdots & \vdots \\ a_{1,2} & a_{2,1} & \ddots & a_{n-1,n-1} & a_{n-1,n} & 1 \\ \vdots & \ddots & a_{n-1,n-1} & 1 & a_{n,n} & 1 \\ a_{1,n} & \dots & a_{n-1,n} & a_{n,n} & 1 & 1 \end{array}\right] $$
For example, a 3 by 3 matrix like this: $$ \left[\begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{array}\right] $$ When we reduce it using reduced row echelon form (RREF), we get: $$ \left[\begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ This system therefore has an infinite number of solutions, it is consistent. You can try it out with any other binary symmetrical matrix with a diagonal of ones and you'll always find at least one solution.
If you have a proof, an idea, or a suggestion on how to proceed, please let me know. Any help is appreciated.