Given $A$ and $B$, find matrix $C$ such that $CA=B$ Hoffman and Kunze, Linear Algebra, ch. 1
$A=\left[\begin{matrix} 1 & -1 \\ 2 & 2 \\ 1 & 0 \end{matrix}\right], B=\left[\begin{matrix} 3 & 1 \\ -4 & 4  \end{matrix}\right]$
Is there a matrix $C$ such that $CA=B$?
I assumed $C=\left[\begin{matrix} a & b & c\\ d & e & f  \end{matrix}\right]$, then computed $CA$, equated each of the four elements of $CA$ with the corresponding elements of $B$, there are many solutions and got one of them.
My question: is there any other method to find C that involves elementary matrices ?
 A: Guide:
$$CA=B$$
$$A^TC^T=B^T$$
We can now perform row operations on augmented matrix $[A^T|B^T]$ to solve for $C^T$.
A: We have
$$
B = C A
$$
where $C$ consists of two unknown row vectors. I am more used to having the unknows on the right hand side ($Ax=b$) so I would, like Siong Thye Goh did, look at the transposed equations:
$$
B^T = (b_1^T, b_2^T)= A^T C^T = A^T (c_1^T, c_2^T)
$$ 
which can be interpreted as two linear systems
$$
A^T c_1^T = b_1^T \\
A^T c_2^T = b_2^T
$$
with column vectors $c_i^T, b_i^T$, which can be solved like usual.
Example: The first system can be solved like
$$
\begin{pmatrix}
1 & 2 & 1 \\
-1 & 2 & 0 
\end{pmatrix}
\begin{pmatrix}
a \\
b \\
c
\end{pmatrix}
=
\begin{pmatrix}
3 \\
1
\end{pmatrix}
\iff \\
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
-1 & 2 & 0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 4 & 1 & 4
\end{array}
\right]
\to 
\\
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & 1 & 3 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\to \\
\left[
\begin{array}{rrr|r}
1 & 0 & 1/2 & 1 \\
0 & 1 & 1/4 & 1
\end{array}
\right]
\iff \\
(a, b, c) = (1-(1/2) c, 1-(1/4) c, c)
$$
where we used $c$ as parameter for the line (1D space) of solutions of that system.
Doing the second calculation, or doing both at the same time, we get:
$$
\left[
\begin{array}{rrr|rr}
1 & 2 & 1 & 3 & -4\\
-1 & 2 & 0 & 1 & 4
\end{array}
\right]
\to
\left[
\begin{array}{rrr|rr}
1 & 0 & 1/2 & 1 & -4\\
0 & 1 & 1/4 & 1 & 0
\end{array}
\right]
\iff \\
(a, b, c) = (1-(1/2) c, 1-(1/4) c, c) \\
(d, e, f) = (-4-(1/2) f, -(1/4) f, f)
$$
for $c, f \in \mathbb{R}$.
