For the following vectors $v_1 = (3,2,0)$ and $v_2 = (3,2,1)$, find a third vector $v_3 = (x,y,z)$ which together build a base for $\mathbb{R}^3$.
My thoughts:
So the following must hold:
$\left(\begin{matrix} 3 & 3 & x \\ 2 & 2 & y \\ 0 & 1 & z \end{matrix}\right) \left(\begin{matrix} {\lambda}_1 \\ {\lambda}_2 \\ {\lambda}_3 \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $
The gauss reduction gives
$ \left(\begin{matrix} 3 & 3 & x \\ 0 & 1 & z \\ 0 & 0 & -\frac{2}{3}x+y \end{matrix}\right) $
(but here I'm not sure if I'm allowed to swap the $y$ and $z$ axes)
For ${\lambda}_1 = {\lambda}_2 = {\lambda}_3 = 0$, this gives me
$ x = 0 \\ y = 0 \\ z = 0 $
Is this third vector $v_3$ building a base of $\mathbb{R}^3$ together with the other two vectors? If not, where are my mistakes?