TITLE: Why does the inverse of a mobius transform not get divided by the determinant?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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
$$ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}, \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. $$