I have to explain all possible solutions of the linear system in Z5:
$$ \begin{cases} 2x+y+3z+t & =3\\ 2x+4z+2t & =0\\ 2y+3z+t & =1 \end{cases} $$
ok now I go with Gauss-Jordan method:
$$ \begin{bmatrix} 2 & 1 & 3 & 1 & 3 \\ 2 & 0 & 4 & 2 & 0 \\ 0 & 2 & 3 & 1 & 1 \\ \end{bmatrix} $$
now with elementary operations on the lines I try to reduce to scale : L2= L2-L1 (L = line)
and we are :
$$ \begin{bmatrix} 2 & 1 & 3 & 1 & 3 \\ 0 & 4 & 1 & 1 & 2 \\ 0 & 2 & 3 & 1 & 1 \\ \end{bmatrix} $$
now L3 = L3 + 2*L2
$$ \begin{bmatrix} 2 & 1 & 3 & 1 & 3 \\ 0 & 4 & 1 & 1 & 2 \\ 0 & 0 & 0 & 3 & 0 \\ \end{bmatrix} $$
now we have a matrix scale reduced.
Rank of matrix is 3(we have 3 pivot: 2,4,3) If i delete last column of "known terms" .. Rank of matrix is in any case 3.
So we should have 5 solutions
The new linear system is :
$$ \begin{cases} 2x+y+3z+t & =3\\ 4y+z+t & =2\\ 3t & =0 \end{cases} $$
...
now I continue with the method of substitution :
$$ \begin{cases} y&= 3x + 2z +3\\ z&= y+2\\ t & =0 \end{cases} $$
$$ \begin{cases} y&= 3x + 2y +4 +3\\ z&= y+2\\ t & =0 \end{cases} $$
$$ \begin{cases} 4y&= 3x + 2\\ z&= y+2\\ t & =0 \end{cases} $$ multiply by 4 first equation $$ \begin{cases} y&= 2x+3\\ z&= y+2\\ t & =0 \end{cases} $$
I brig the "x" to the left in first equation :
$$ \begin{cases} 2x &= y+2\\ z&= y+2\\ t & =0 \end{cases} $$
multiply by 3 first equation :
$$ \begin{cases} x &= 3y+1\\ z&= y+2\\ t & =0 \end{cases} $$
and afterwards I don't know how to continue for explain the solutions..
help me, how can I continue for explain solution?!