<s>[INST] Find nullspace of this matrix The question I have is this:
1) Find a basis for the solution space of the homogeneous linear system.
$$x - y + 2z = 0 \\
2x + y = 0 \\
x − 4y + 6z = 0$$
So I convert to a matrix:
$$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & 0 \\ 1 & -4 & 6 \end{bmatrix}$$
$$ -> \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -4 \\ 0 & -3 & 4 \end{bmatrix}$$
$$ -> \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -4 \\ 0 & 0 & 0 \end{bmatrix}$$
So is this right?
Let $z = t, y = \frac{4}{3}t, x = -2t + \frac{4}{3}t$
so:
$$ -> \begin{bmatrix} x & y & z \end{bmatrix}$$
$$ -> \begin{bmatrix} \frac{-2}{3}t & \frac{4}{3}t & t \end{bmatrix}$$
$$ -> t * \begin{bmatrix} \frac{-2}{3} & \frac{4}{3} & 1 \end{bmatrix}$$
So the basis is:
$$ -> \begin{bmatrix} \frac{-2}{3} & \frac{4}{3} & 1 \end{bmatrix}$$
Is this right?
More generally, the reason we do it this way is because we're trying to solve for x in the equation $Ax = 0$ so we have to reduce A, setup a system of equations that result when you multiply A by X, setup any free variables, extract the coefficients and that's the basis of the nullspace (the set of all solutions to Ax = 0). Does that sound right?
 [/INST] Looks to me like the right method to get the rational basis for the null space. Quick check also shows that 
$$
\left[
    \begin{matrix}
    1 & -1 & 2 \\
    2 & 1 & 0 \\
    1 & -4 & 6 \\
    \end{matrix}
\right]
\left[
    \begin{matrix}
    -2/3  \\
    4/3  \\
    1  \\
    \end{matrix}
\right] =
\left[
    \begin{matrix}
    0  \\
    0  \\
    0  \\
    \end{matrix}
\right]
$$
</s>