Carregando a matriz "m" com a entrada do sistema.

A solução proposta está em "s".
Como o sistema escalonado tem duas linhas nulas, faremos as variáveis
(que podemos escolher o valor) como x=0 e y=0, de forma
que a solução "s" acaba com "0, 0".

Multiplicamos "m . s" e guardamos em "v" 

O objetivo é que "v=v0", onde "v0 = [A,B,C,...,X,Y]" (transposto).
Como isso não aconteceu, podemos usar as equações de 
compatibilidade (correspondentes às duas linhas nulas do escalonamento)
para somar a "v". Chamaremos esse ajuste de "aj".
No final, substituimos 2e 4 por 0, 3 e 5 por 1.
Conseguimos finalmente ver que mod(v+aj,2)=v0.

Avalie todas as células de uma vez com Shift+Ctrl+R e leia até o final.

m: matrix(
[1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,1,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0,0,1,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,0,0,0,0,1],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1,0],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1,1],
[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1,1])$

s: matrix(
[B+C+D+H+J+N+O+T],
[A+B+D+E+G+M+N+O+S],
[A+C+D+E+F+H+I+M+N+P+Q+R+S+T+V],
[A+D+F+G+H+J+K+L+N+O+P+Q+R+T+U+X],
[B+C+E+F+K+M+O+R+T+U+V],
[C+E+G+H+J+M+S+T],
[B+D+F+G+I+J+N+P+Q+R+V],
[A+B+D+I+J+K+L+N+U+V+X],
[B+D+F+G+I+J+L+R+S+T+X],
[A+B+C+D+G+H+J+L+M+N+P+Q+V+X],
[E+I+J+M+O+P+Q+R+S+V],
[B+C+D+F+H+J+K+L+N+O+P+Q+R+T+U+X],
[B+C+E+F+J+K+M+N+R+U+V],
[A+B+C+G+I+M+N+O+S],
[A+B+E+H+I+J+K+N+O+P+Q+S+U],
[B+D+F+G+I+J+L+N+U+V+X],
[B+F+G+H+N+P+Q+S+T+V+X],
[C+E+G+H+J+K+M+P+Q+S+T+U],
[D+H+I+J+L+P+Q+S+T+V+X],
[A+B+D+E+K+L+N+O+U+V+X],
[D+E+H+L+M+O+P+R+T+U+V],
[B+E+F+G+H+I+J+M+N+O+P+R+T+U+X],
[D+H+I+J+L+P+Q+S+T+V],
[0],
[0])$


BC: B+C+D+F+H+J+K+L+N+O+P+R+T+V+W+X$
AB: A+B+D+E+K+L+N+O+U+V+X+Y$

v: transpose(m . s)$

As variáveis BC e AB são nulas por serem as equações de
compatibilidade do sistema escalonado (duas linhas nulas no final).
O vetor aj é 0, pois AB=BC=0 pelas equações de compatibilidade.
Ele será usado para mostrar que a solução obtida resolve.

aj: matrix([0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,BC,0,AB])$
subst([2=0,4=0,3=1,5=1],v+aj);

Se a última saida for [A,B,C,...,W,X,Y]=v0, conseguimos, pois o sistema original é
                   m . x = v0 (módulo 2)