Finding the basis of a column space I was given the following question:

Assume that $A$ is row equivalent to $B$. Find bases for Null A and Col A.  $$A=\left[\begin{matrix}-2&6&-2&-6\\2&-9&-6&2\\-3&12&5&-5\\\end{matrix}\right],B=\left[\begin{matrix}1&0&9&7\\0&3&8&4\\0&0&0&0\\\end{matrix}\right]$$

I know that to find the basis of the null space, I can set $A$ equal to zero and use the variables from the free columns. I also know that to find the basis of the column space I can just use all the rows with leading ones. My question involves the column space. In order to minimize this down into REF, do I put A and B into an augmented matrix? Or can I find the column space without using my B at all - by setting to zero in the same way as I do for my null space?
Edit:
Right now, I created an augmented matrix, and found the REF. My columnspace became: $$\left(\begin{matrix}1\\0\\0\\\end{matrix}\right),\left(\begin{matrix}-4\\1\\0\\\end{matrix}\right),\left(\begin{matrix}0\\\frac{-1}{2}\\1\\\end{matrix}\right)$$Is this correct?

Doing row operations on $A$ doesn't change which columns are linearly independent.
In $B$, you can see that the first two columns are independent, and all others are linear combinations of those; more generally, in a RREF, if you look for the first nonzero entry in each row and circle those (in this case, that's entry $(1,1)$ and $(2,2)$) then the COLUMNS containing circled elements span the column space.
What about the null space? For each column NOT containing a circled entry (in this case columns 3 and 4), you can make the corresponding entries all zeroes except for a single $1$, so you create vectors
$$
\pmatrix{\star\\ \star \\ 0 \\ 1}, \pmatrix{\star\\ \star \\ 1 \\ 0}
$$
in your example; then you ask "what values must I put in to replace the stars to make these vectors satisfy $Ax = 0$? It'll turn out to be easy to answer that (from the row-echelon form), and that'll give you a basis for the nullspace.
