Originally Posted by
TheEmptySet You can paramterize the soutuions as follows
let $\displaystyle z=t$
Then the equation above the row of zero's is
$\displaystyle -5y+10z=-15 \iff -5y+10(t)=-15 \iff y=3+2t$ and using these we can solve for x
$\displaystyle x+2(3+2t)-4(t)=10 \iff x+6+4t-4t=10 \iff x=4$
So finally we get the solution
$\displaystyle (4,3+2t,t)$
This can be broken up into the particular and complimentry solution
$\displaystyle (4,3+2t,t)=(4,3,0)+t(0,2,1)$
where $\displaystyle y_p=(4,3,0) \mbox{ and } y_c=t(0,2,1)$
This is true for any $\displaystyle t \in \mathbb{R}$
Note that $\displaystyle y_c$ spans the null space of the homogenious system $\displaystyle Ax=0$