Really tried to format it on here, but it looked horrible...
I need help understanding what exactly The boxed section of the question is asking. I understand how to find B and it makes sense, but obviously the same method wont work for finding the matrix representation for C. Any help would be greatly appreciated.
Thanks.
Writing a matrix for a linear transformation, requires us to assume or specify a priori, a choice of basis, on both the domain space as well as the codomain space. Once this is done, we can write the matrix down with respect to these bases. Here we have $T : \mathbb R^3 \to \mathbb R^2$ but we will specify a more general setting.
Let $S : V \to W$ be a linear transformation, and let $\{a_1,...,a_n\}$ and $\{b_1,...,b_m\}$ be bases for $V$ and $W$ respectively. This is how you would write down the matrix of $T$. It will have $m \times n$ dimensions.
Now, in part $(a)$, no bases are specified. This usually means that on both vector spaces the canonical basis $e_i$ ($1$ in the $i$th position and zero elsewhere) is put, and then we write the matrix with respect to these bases. You have done this correctly.
In part $(b)$, a basis for the domain is specified. So the canonical basis for codomain remains. That is, just find $T(b_1)$, get the coefficients of $x,y,z$ out in to get the first column, and then put all the $T(b_i)$ together to get the matrix.
In part $(c)$, a basis for the codomain is specified, so you have to now assume that we have the canonical basis for the domain space. So what you have to do is this : find $T(e_1)$, where $e_1 = (1,0,0)$, and find the unique coefficients so that $T(e_1) = c_1(1,2) + c_2(1,3)$. Then, $[c_1,c_2]^T$ is the first column of $T$. Do the same with $e_2$ and $e_3$ to get the other columns , and put them together to get the answer.