Why does the inverse of a mobius transform not get divided by the determinant? for a mobius transform $\frac{az+b}{cz+d}$ it's inverse is given by $\frac{dw-b}{-cw+a}$ since this can be put into a matrix form why is there not a requirement for the inverse to be divided by the determinant?
 A: Short answer: For any given mobius transformation, there​ are many different matrices that represent it. Once you realise this, you will see that we can just ignore the term you're asking about.
Slightly longer answer: For any non-zero complex number $t$, we have
$$
\frac{az+b}{cz+d}=\frac{atz+bt}{ctz+dt}
$$
This means that if you calculate the inverse of the matrix of the coefficients, the $\frac1{\det}$ term may be cancelled away, not because of some matrix property, but because the mobius transformation is given, in the end, by a fraction and not a matrix.
Ultimate answer: Forget the matrices. Until you have this completely worked out and fully understood, the matrices distract you from what's really going on. Once you have this completely worked out and fully understood, the matrices are not needed. Calculate the inverse by hand, directly from the original fraction, or insert one expression into the other, and see that it works out.
A: There is some ambiguity in representing a Mobius transformation by a matrix. Since we have
$$ \frac{az + b}{cz + d} = \frac{e(az + b)}{e(cz + d)} = \frac{(ea)z + (eb)}{(ec)z + (ed)}  $$
for any $e \neq 0$, the matrices
$$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, A' = \begin{pmatrix} ea & eb \\ ec & ed \end{pmatrix} = eA $$
represent the same Mobius transformation. The inverse Mobius transformation is then represented by both the matrices
$$ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. $$
A: The correspondence between the Möbius transform and the matrix is not one-to-one.
The quadruplets (or the corresponding matrices) $(a,b,c,d)$ and $(\lambda a,\lambda b,\lambda c,\lambda d)$ correspond to the same transform for any $\lambda\neq0$.
Therefore you are free to multiply the coefficients with any number (as long as the number is the same for each coefficient, of course).
In the formula of the inverse transform you could divide each coefficient by the determinant $ad-bc$.
But this would not change the transform.
Notice that you do not want to divide the formula, you want to divide each coefficient; otherwise it does not correspond to dividing the matrix by the determinant.
Because of this scaling invariant, you could in fact assume $ad-bc=1$.
All Möbius transforms can be written in this form because of the scaling invariance.
