Finding a matrix associated with a given domain and target space of a transformation For the following matrix $A$, first find a nonzero vector $X$ in the nullspace of $A$.Then compute the matrix of $T_A$using the ordered basis $\mathcal{B}=\{[1, 0]^t,X\}$ for the domain and the ordered basis $\bar{\mathcal{B}}=\{A_1, [0, 1, 0]^t, [0, 0, 1]^t\}$ for the target space where $A_1$ is the first column of $A$.
$$A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{bmatrix}$$
I was able to find the vector $X$ as $\begin{bmatrix} -2 \\ 1 \end{bmatrix}$, and thus obtained that the ordered basis $\mathcal{B}$ is given by $$\mathcal{B} = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -2 \\ 1 \end{bmatrix} \right\}$$
and $$\bar{\mathcal{B}} = \left\{ \begin{bmatrix} 1 \\ 2\\ 3 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\}$$
but am unable to continue. I believe that the transformation $T_A: \mathbb{R}^2 \rightarrow \mathbb{R}^3$, but I don't know how to find the matrix that corresponds to this.
 A: To find the matrix corresponding to linear transformation A from X to Y, in a given basis for X and a given basis for Y, apply A to each basis vector for X then write the result as a linear combination of the basis vectors for Y. The coefficients form a column in the matrix. 
Here the basis vectors for X are $u= \begin{bmatrix}1 \\ 0 \end{bmatrix}$ and $v= \begin{bmatrix}1 \\ -2\end{bmatrix}$.  $Au= \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$ and $Av= \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ 3 & 6\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}-3 \\ -6 \\ -9 \end{bmatrix}$. 
Write those in terms of the basis vectors for Y, $p= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$, $q= \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$, and $\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}$.
So we want to find A, B, and C so that $A\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ B\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}+ C\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}A \\ 2A+ B \\ 3A+ C \end{bmatrix}= \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}$ which is easily solved: A= 1, B= C= 0. 
And we want to find A, B, and C so that $A\begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}+ B\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}+ C\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix}A \\ 2A+ B \\ 3A+ C \end{bmatrix}= \begin{bmatrix}-3 \\ -6 \\ -9 \end{bmatrix}$.  That is almost as easy to solve: A= -3 so 2A+ B= -6+ B= -6 so B= 0 and 3A+ C= -9+ C= -9.  C= 0.  A= -3 and B= C= 0.
Using those as columns the matrix we want is $\begin{bmatrix}1 & -3 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$
