I have the following system of equations:
$$\left\{\begin{array}{c} x+2y-2z+2s-t=0\\ x+2y-z+3s-2t=0\\ 2x+4y-7z+s+t=0 \end{array}\right.$$
Which forms the following matrix
$$\left[\begin{array}{ccccc|c} 1 & 2 & -2 & 2 & -1 & 0\\ 1 & 2 & -1 & 3 & -2 & 0\\ 2 & 4 & -7 & 1 & 1 & 0 \end{array}\right]$$
Which I then row reduced to the following form:
$$\left[\begin{array}{ccccc|c} 1 & 2 & 0 & 4 & -3 & 0\\ 0 & 0 & 1 & 1 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right]$$
I am unsure from this point how to find the basis for the solution set. Any help of direction would be appreciated. I know that the dimension would be $n-r$ where $n$ is the number of unknowns and $r$ is the rank of the matrix but I do not know how to find the basis.
First solve the system, assigning parameters to the variables which correspond to non-leading (non-pivot) columns: $$\eqalign{ &t=\alpha\cr &s=\beta\cr z+s-t=0\quad\Rightarrow\quad &z=\alpha-\beta\cr &y=\gamma\cr x+2y+4s-3t=0\quad\Rightarrow\quad &x=3\alpha-4\beta-2\gamma\ .\cr}$$ So the solution set is $$\left\{\pmatrix{3\alpha-4\beta-2\gamma\cr \gamma\cr \alpha-\beta\cr \beta\cr \alpha\cr}\ \Bigg|\ \alpha,\beta,\gamma\in{\Bbb R}\right\}$$ which can be written $$\left\{\alpha\pmatrix{3\cr0\cr1\cr0\cr1\cr}+\beta\pmatrix{-4\cr0\cr-1\cr1\cr0\cr}+\gamma\pmatrix{-2\cr1\cr0\cr0\cr0\cr}\ \Bigg|\ \alpha,\beta,\gamma\in{\Bbb R}\right\}\ .$$ The three vectors shown span the solution set; it is also not too hard to prove that they are linearly independent; therefore they form a basis for the solution set.