Let $\bar{y} = \frac{A\bar{x} +\bar{b}}{\bar{c}^{T}\bar{x} + d}$, where $A$ is $n \times m$ matrix with $n, m \in \mathbb R_{+}$. Let $f(x) := \bar{y}$ so $f : \mathbb R^{m} \mapsto \mathbb R^{n}$. The denominator and numerator are convex and affine so $\bar{y}$ must be convex and affine because $\bar{y}$ is essentially a perspective function with affine and convex functions. $d \in \mathbb R$ is scalar.
I want to calculate its inverse.
Switch $\bar{x}$ to $\bar{y}^{-1}$ and $\bar{y}$ to $\bar{x}$ and solve.
$\bar{x}(\bar{c}^{T}\bar{y}^{-1}+d) - A\bar{y}^{-1} - \bar{b} = 0$
$(\bar{x} \bar{c}^{T} - A ) \bar{y}^{-1} + (\bar{x} d - \bar{b} ) = 0$
So
$\bar{y}^{-1} = (\bar{x} \bar{c}^{T} -A)^{-1}(\bar{b} - \bar{x} d)$
if $(\bar{x} \bar{c}^{T} -A)$ is invertible.
Alert:
$\bar{x}(\bar{c}^{T}\bar{y}^{-1}) = (\bar{x}\bar{c}^{T})\bar{y}^{-1}$ where $c^{T}\bar{y}^{-1}$ is contant but the dimensions with $\bar{x}\bar{c}^{T}$ may not match. Something very wrong here, any rescue? Evidently I should have certain assumptions about the dimensions so that the operations above are correct. Is there some easy way to find the inverse mapping for the function $\bar{y}$?