# difficult systems of equations

• September 12th 2007, 11:06 PM
lord12
difficult systems of equations
<0,0,0,1> <0,0,2,1> <0,3,2,1> <0,1,-1,1> are vectors

w = 0
2z + w = 0
3y + 2z +w = 0
y - w +z=0

How do I find other solutions besides w=z=y=x=0 since I know these vectors are linearly dependent.
• September 12th 2007, 11:48 PM
Opalg
What makes you think that x=0?
• September 12th 2007, 11:50 PM
tukeywilliams
Yes $x$ is a free variable.

So $x = t$ (where $t$ is a varying parameter)

$y = z = w = 0$.