I'm asking this to see if my answer is correct. let $S$ be a subspace of $\mathbb{R}^5$ made up of the vectors
$x = [x_1,x_2,x_3,x_4,x_5]^T$
using the equations
$(x_1)+6(x_3)-2(x_4)-3(x_5)=0$
$(x_2)+3(x_3)+(x_4)+2(x_5)=0$
find a basis for $S$
in my answer i got the solution for $x_1$ and $x_2$
$(x_1)=-6(x_3)+2(x_4)+3(x_5)$
$(x_2)=-3(x_3)-(x_4)-2(x_5)$
$$ [x_1,x_2,x_3,x_4,x_5]^T = x_3[-6,-3,1,0,0]^T, x_4[2,-1,0,1,0]^T, x_5[3,-2,0,0,1]^T $$ Basis of $S = [-6,-3,1,0,0]^T,[2,-1,0,1,0]^T,[3,-2,0,0,1]^T$
Are these three vectors correct for the basis of the subspace?