So I have this linear system:
$ \begin{align*} -u + v &= y_1\\ u + v &= y_2 \\ 2u + v &= y_3 \end{align*} $
After doing gaussian elimination, I get:
$ \begin{align*} u = y_3 - y_2 \\ v = 2y_2 - y_3 \\ y_1 -3y_2 + 2y_3 = 0 \end{align*} $
I'm asked to find for what values of $y_1, y_2, y_3$ this system has a solution. The answer to that would be:
$\left[ \begin{array}{ccc} 1 & -3 & 2 \\ \end{array} \right] * \left[ \begin{array}{ccc} y_1 \\ y_2 \\ y_3 \end{array} \right] = 0$
i.e. the null space. How do I find a basis for the null space though?