Determining the matrix representations of functions. Show that each function\begin{align}f(x)=\dfrac{ax+b}{cx+d}, g(x)=\dfrac{ex+f}{jx+h}\end{align}
can be represented by matrices such that the matrix representation of the composition $g(f(x))$ is the product of the matrices of $g$ and $f$.
I tried expressing $\frac{ax+b}{cx+d}$ to $\frac{a}{c}+\frac{bc-da}{c(cx+d)}=ac^2x+bc+cda-da$. Which made me think might $x$ be a constant, vector, etc...? I would like to as for hints on this.
 A: You need to know some group theory up to the first isomorphism theorem to understand the matrix representation.
The map from $\mathrm{SL}(2,\Bbb F)$ to the group of all Möbius transformations, sending $\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ to the Möbius transformation $f$, where $f(x)=\frac{ax+b}{cx+d}$, is a surjective group homomorphism, with the kernel of this map being $\{I,-I\}$. So by the first isomorphism theorem, $\mathrm{SL}(2,\Bbb F)/\{I,-I\}\cong$ Möbius transformation group. The induced isomorphism tells us that $\begin{bmatrix}a & b\\ c & d\end{bmatrix}\{I,-I\}$ represents the function $f$. But we may drop $\{I,-I\}$ and simply say that $\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ represents $f$.
For details about the above claims with proofs, consult Alan Beardon's Algebra and Geometry, the chapter on Möbius transformations.
A: Hint:  Try evaluating $g(f(x))$.
By the way, in general it is not true that
$$\frac{a}{c}+\frac{bc-da}{c(cx+d)} = ac^2 x + bc + cda - da.$$
As hinted in the comments by Ethan, try searching for Möbius Transformation.
