show that there exists a function $g(x)=\frac{\alpha x+\beta}{\gamma x+\delta}$ such that $f(g(x))=x$ question: Let $a,b,c,d$ be given constants with the property that $ad-bc\neq0$. If $f(x)=\frac{ax+b}{cx+d}$, show that there exists a function $g(x)=\frac{\alpha x+\beta}{\gamma x+\delta}$ such that $f(g(x))=x$. Also show that for these two functions it is true that $f(g(x))=g(f(x))$.
how can I get the answer $\alpha=\frac{d}{ad-bc}$, $\beta=\frac{-b}{ad-bc}$, $\gamma=\frac{-c}{ad-bc}$, $\delta=\frac{a}{ad-bc}$
my approach: first calculate the value of $f(g(x))$
$$f(g(x))=\frac{(a\alpha+b\gamma)x+(a\beta+b\delta)}{(c\alpha+d\gamma)x+(c\beta+d\delta)}$$
then, $f(g(x))=x$
$$
\begin{align}
\frac{(a\alpha+b\gamma)x+(a\beta+b\delta)}{(c\alpha+d\gamma)x+(c\beta+d\delta)}&=x\\
(a\alpha+b\gamma)x+(a\beta+b\delta)&=(c\alpha+d\gamma)x^2+(c\beta+d\delta)x
\end{align}
$$
and I stuck at this step. please help!
 A: You're trying to find the inverse function of $f(x)$, so set the function equal to $y$ and solve for $x$:
$$y = \frac{ax+b}{cx+d}$$
$$ycx+yd = ax+b$$
$$ycx-ax = b-yd$$
$$x(yc-a) = b-yd$$
$$x = \frac{-dy+b}{-a+cy}=\frac{dy-b}{a-cy}.$$
Interchange $x$ and $y$ and you have
$$g(x) =\frac{dx-b}{a-cx}.$$
If you divide top and bottom by $ad-bc$, you'll have the answer you want.
A: $g$ is the Moebius transformation corresponding to the nonsingular matrix. Then $f$ corresponds to the inverse matrix.
A: $f$ is a Möbius transformation. In B. Goddard's answer an inverse has been computed.
However, there are two problems.

*

*You did not specify domain and range of $f$. Usually $\alpha, \beta, \gamma, \delta$ are understood as complex numbers and $x$ is understood as a complex variable. But on principal everything could be real, rational or something else. Anyway, let us assume that we work in $\mathbb C$.


*Unless $\gamma = 0$, your function $f$ is not a function $\mathbb C \to \mathbb C$. In fact, it is undefined for $x = -\frac \delta \gamma$, thus we have a function $f : \mathbb C \setminus \{ -\frac \delta \gamma \} \to \mathbb C$. Unfortunalely it is not a bijection. Look at the "inverse" $g(x)  = \frac{\delta x - \beta}{\alpha - \gamma x}$. This is undefined for $x = \frac \alpha \gamma$, hence $\frac \alpha \gamma$ is not in the image of $f$ (you can also check directly that $f(x) = \frac \alpha \gamma$ does not have a solution). Thus we have to consider
$$f : \mathbb C \setminus \{ -\frac \delta \gamma \} \to \mathbb C \setminus \{ \frac \alpha \gamma \} .$$
Then in fact an inverse is given by
$$g : \mathbb C \setminus \{ \frac \alpha \gamma \} \to \mathbb C \setminus \{ -\frac \delta \gamma \}.$$
An alternative approach is to consider the extended complex plane (Riemann sphere) $\mathbb  C^* = \mathbb C \cup \{\infty\}$. Then $f$ extends to a function $f^* : \mathbb  C^* \to \mathbb  C^*$ by setting $f^*(x) = f(x)$ for $x \in \mathbb C \setminus \{ -\frac \delta \gamma \}$, $f^*( -\frac \delta \gamma ) = \infty$ and $f^*(\infty) = \frac \alpha \gamma$. You can similarly extend $g$ to $g^* : \mathbb  C^* \to \mathbb  C^*$. Then $g^*$ is the inverse of $f^*$.
