How to determine if a set spans a vector space $\lbrace (2, -6), (-1,4), (-3, 9)\rbrace$
I row reduced this to get the rows $(2,-1,-3); (0,1,0)$. I want to check if this spans $\mathbb{R}^2$. I don't see exactly what to call the free variables. If I use $c_1, c_2, c_3$ for each column respectively then I get $c_2= b$ and $2c_1 - c_2 -3c_3 = a$. $c_3$ is always allowed to be anything. What decides if I have two free variables? 
 A: We know that the set $S=\lbrace \langle 2,-6\rangle,\langle -1,4\rangle,\langle -3,9\rangle\rbrace$ spans $\mathbb{R}^2$ if there are exactly two linearly independent vectors in $S$. Observe that $$-\frac{3}{2}\cdot\langle2,-6\rangle=\langle-3,9\rangle$$
so these two vectors are not linearly independent. To test if $\langle -1,4\rangle$ is independent of the other two vectors, evaluate 
$$\text{RREF}\begin{pmatrix}
-1 & 4 \\
2 & -6
\end{pmatrix}=\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}=I_2$$
Because the matrix formed by the two vectors is row equivalent to the $2\times 2$ identity matrix, they are linearly independent. Therefore, $S$ spans $\mathbb{R}^2$. 
A: The trick to putting the vectors in a matrix is that row reduction preserves linear independence of the columns.   So to determine which of the original columns are independent,  just check which columns in the reduced matrix are.  Thus, for instance, since columns $1$ and $2$ in the reduced matrix are independent,  so are $1$ and $2$ from the original vectors.
But, it should be noted that since you only need $2$ vectors to span $\mathbb R^2$, and since checking if two vectors are independent just means checking if they are multiples of each other, this trick is barely necessary here...
Addressing your work at the end, use $c_1,c_2,c_3$ for the coordinates (clearly what you meant). Then $c_2=b$, if you choose a specific $(a,b)$ in the image. Then note that $2c_1-b-3c_3=a$.  So choose $c_3$ freely and then $c_1$ is determined.   Or vice versa (choose any $c_1$ and then $c_3$ is determined). So, the solution space for a given $(a,b)$ is only $1$-dimensional (but that there's a solution at all is what we're interested in). Incidentally,  $1$ is also the nullity.
