Why doing Möbius transformation behaves like operating a 2X2 matrix of coefficients of Möbius transformation? So, I was doing proof that inverse of Möbius transformation is again a Möbius transformation.
I started with $w= f(z)=\frac{az+b}{cz+d}$. To find the inverse, I found z in terms of w by doing simple algebra and I got $z=\frac{dw-b}{-cw+a}$=g(w) (say). Alternatively, I can just look at the inverse of the matrix of coefficients of f(z) i.e. \begin{pmatrix}a&b\\c&d\end{pmatrix} and find its inverse to get $\frac{1}{ad-bc} \begin{pmatrix}d&-b\\-c&a\end{pmatrix}$. This inverse matrix has elements corresponding to the inverse transformation. I just can't see why. Please explain without using any group theory if possible.

The most natural action of $2\times 2$ matrices is on 2-dimensional vectors:
$$\tag1 \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\cx+dy\end{pmatrix}.$$
And here, we naturally have $A(Bv)=(AB)v$, i.e., composition of maps is the same as first multiplying matrices.
Now noe that a scalar multiple of a vector is mapped to the same multiple of the image vector, i.e., $A(tv)=tAv$. Hence we can view the map $(1)$ also as acting on something that does not change when we replace $x$ and $y$ by $tx$ and $ty$, respectively. The obvious choice for such a something is the quotient $\frac xy$ (with caveas about the case $y=0$). With this convention, our matrix maps $z:=\frac xy$ to
$$ \frac{ax+by}{cx+dy}=\frac{a\frac xy+b}{c\frac xy+d}=\frac{az+b}{cz+d}.$$
We just invented Möbius transformations! And the composition property of course transfers from the original. 
