Answer
We prove that ${\rm{Jac}}\left( {{G^{ - 1}}} \right) = {\rm{Jac}}{\left( G \right)^{ - 1}}$
Work Step by Step
Let $I$ be the identity map such that $I\left( {u,v} \right) = \left( {u,v} \right)$. So, $x=u$, $y=v$.
Evaluate the Jacobian of $I$:
${\rm{Jac}}\left( I \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right| = 1$
Suppose ${G^{ - 1}}$ is the inverse of the mapping $G$. So, $G \circ {G^{ - 1}} = {G^{ - 1}} \circ G = I$.
Evaluate the Jacobian:
${\rm{Jac}}\left( {G \circ {G^{ - 1}}} \right) = {\rm{Jac}}\left( {{G^{ - 1}} \circ G} \right) = {\rm{Jac}}\left( I \right) = 1$
Using the result in Exercise 50:
${\rm{Jac}}\left( G \right){\rm{Jac}}\left( {{G^{ - 1}}} \right) = 1$
${\rm{Jac}}\left( {{G^{ - 1}}} \right) = \frac{1}{{{\rm{Jac}}\left( G \right)}}$
Hence, ${\rm{Jac}}\left( {{G^{ - 1}}} \right) = {\rm{Jac}}{\left( G \right)^{ - 1}}$.
