Answer
(a) we show that if Eq. (6) holds for ${\cal D}$, it also holds for ${\cal D} + {\bf{u}}$.
(b) we show that Eq. (6) also holds for each triangular half of any rectangle.
(c) using results in part (a) and part (b), we show that Eq. (6) holds for arbitrary parallelograms.
Work Step by Step
Let ${\cal D}$ be any set in the $uv$-plane. Since $G$ is a linear map, by linearity properties, Eq. (1):
$G\left( {{\cal D} + {\bf{u}}} \right) = G\left( {\cal D} \right) + G\left( {\bf{u}} \right)$
This implies that $G$ maps ${\cal D} + {\bf{u}}$ to the translate $G\left( {\cal D} \right) + G\left( {\bf{u}} \right)$.
Since the shape of ${\cal D}$ does not change by translating to ${\cal D} + {\bf{u}}$, we have
(1) ${\ \ \ \ \ }$ ${\rm{Area}}\left( {\cal D} \right) = {\rm{Area}}\left( {{\cal D} + {\bf{u}}} \right)$
Similarly, the shape of $G\left( {\cal D} \right)$ does not change by the translate $G\left( {\cal D} \right) + G\left( {\bf{u}} \right)$, so we have
(2) ${\ \ \ \ \ }$ ${\rm{Area}}\left( {G\left( {\cal D} \right)} \right) = {\rm{Area}}\left( {G\left( {{\cal D} + {\bf{u}}} \right)} \right)$
If Eq. (6) holds for ${\cal D}$, then
${\rm{Area}}\left( {G\left( {\cal D} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {\cal D} \right)$
Using equation (1) and (2), we obtain
${\rm{Area}}\left( {G\left( {{\cal D} + {\bf{u}}} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {{\cal D} + {\bf{u}}} \right)$
The last equation implies that Eq. (6) holds for ${\cal D} + {\bf{u}}$.
We conclude that if Eq. (6) holds for ${\cal D}$, it also holds for ${\cal D} + {\bf{u}}$.
(b) Let the rectangle ${\cal D}$ in the $uv$-plane in Figure 19(A) be ${\cal D}:\left[ {0,a} \right] \times \left[ {0,b} \right]$. So, ${\rm{Area}}\left( {\cal D} \right) = ab$.
Since $G$ is a linear map, we can write
$G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
A&C\\
B&D
\end{array}} \right| = AD - BC$
The vertices of ${\cal D}$ are $\left( {0,0} \right)$, $\left( {a,0} \right)$, $\left( {a,b} \right)$, $\left( {0,b} \right)$.
Using $G$ we find the images of these vertices:
$\begin{array}{*{20}{c}}
{\left( {u,v} \right)}&{G \to }&{\left( {x,y} \right)}\\
{\left( {0,0} \right)}&{}&{\left( {0,0} \right)}\\
{\left( {a,0} \right)}&{}&{\left( {aA,aB} \right)}\\
{\left( {a,b} \right)}&{}&{\left( {aA + bC,au + bD} \right)}\\
{\left( {0,b} \right)}&{}&{\left( {bC,bD} \right)}
\end{array}$
Since $G$ is linear, the image of ${\cal D}$ is a parallelogram spanned by $\left( {aA,aB} \right)$ and $\left( {bC,bD} \right)$. Using Eq. (10) in Section 13.4, the area of $G\left( {\cal D} \right)$ is
${\rm{Area}}\left( {G\left( {\cal D} \right)} \right) = \left| {\det \left( {\begin{array}{*{20}{c}}
{aA}&{aB}\\
{bC}&{bD}
\end{array}} \right)} \right| = \left| {abAD - abBC} \right|$
But ${\rm{Area}}\left( {\cal D} \right) = ab$ and ${\rm{Jac}}\left( G \right) = AD - BC$. Therefore,
${\rm{Area}}\left( {G\left( {\cal D} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {\cal D} \right)$
Since $A$, $B$, $C$, $D$ are arbitrary, we conclude that Eq. (6) holds for all rectangles in the $uv$-plane with vertex at the origin and sides parallel to the axes.
Consider now, the triangular region ${{\cal D}_1}$, which is a half of ${\cal D}$, as is shown in Figure 19(A).
Then, we have ${\rm{Area}}\left( {{{\cal D}_1}} \right) = \frac{1}{2}{\rm{Area}}\left( {\cal D} \right)$.
Since $G$ is linear, we get ${\rm{Area}}\left( {G\left( {{{\cal D}_1}} \right)} \right) = \frac{1}{2}{\rm{Area}}\left( {G\left( {\cal D} \right)} \right)$.
Applying Eq. (6) to ${\cal D}$, we obtain
$2{\rm{Area}}\left( {G\left( {{{\cal D}_1}} \right)} \right) = 2\left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {{{\cal D}_1}} \right)$
Or
${\rm{Area}}\left( {G\left( {{{\cal D}_1}} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {{{\cal D}_1}} \right)$
Thus, we conclude that Eq. (6) also holds for each triangular half of such a rectangle.
(c) Referring to Figure 19 (B), let ${\cal R}$ denote the outer rectangle and ${{\cal D}_0}$ denote the shaded parallelogram. Let ${{\cal D}_1},{{\cal D}_2},...,{{\cal D}_6}$ denote other small rectangles and triangles in the interior of ${\cal R}$. So, we have the difference of region:
${{\cal D}_0} = {\cal R} - \mathop \sum \limits_{i = 1}^6 {{\cal D}_i}$
So, the area of ${{\cal D}_0}$:
(3) ${\ \ \ \ \ }$ ${\rm{Area}}\left( {{{\cal D}_0}} \right) = {\rm{Area}}\left( {\cal R} \right) - \mathop \sum \limits_{i = 1}^6 {\rm{Area}}\left( {{{\cal D}_i}} \right)$
Under the linear mapping $G$, we obtain the image of ${{\cal D}_0}$:
$G\left( {{{\cal D}_0}} \right) = G\left( {{\cal R} - \mathop \sum \limits_{i = 1}^6 {{\cal D}_i}} \right)$
Since $G$ is linear, we get
$G\left( {{{\cal D}_0}} \right) = G\left( {\cal R} \right) - \mathop \sum \limits_{i = 1}^6 G\left( {{{\cal D}_i}} \right)$
Thus, the area:
(4) ${\ \ \ \ \ }$ ${\rm{Area}}\left( {G\left( {{{\cal D}_0}} \right)} \right) = {\rm{Area}}\left( {G\left( {\cal R} \right)} \right) - \mathop \sum \limits_{i = 1}^6 {\rm{Area}}\left( {G\left( {{{\cal D}_i}} \right)} \right)$
From part (a) and part (b), we know that Eq. (6) holds for rectangles and triangles. So, applying Eq. (6) to ${\cal R}$ and ${{\cal D}_i}$ in equation (4) gives
${\rm{Area}}\left( {G\left( {{{\cal D}_0}} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {\cal R} \right) - \mathop \sum \limits_{i = 1}^6 \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {{{\cal D}_i}} \right)$
${\rm{Area}}\left( {G\left( {{{\cal D}_0}} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|\left( {{\rm{Area}}\left( {\cal R} \right) - \mathop \sum \limits_{i = 1}^6 {\rm{Area}}\left( {G\left( {{{\cal D}_i}} \right)} \right)} \right)$
By equation (3), ${\rm{Area}}\left( {{{\cal D}_0}} \right) = {\rm{Area}}\left( {\cal R} \right) - \mathop \sum \limits_{i = 1}^6 {\rm{Area}}\left( {{{\cal D}_i}} \right)$. Therefore,
${\rm{Area}}\left( {G\left( {{{\cal D}_0}} \right)} \right) = \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{Area}}\left( {{{\cal D}_0}} \right)$
Thus, Eq. (6) holds for the parallelogram ${{\cal D}_0}$.
Since ${{\cal D}_0}$ is arbitrary and we can always construct any number of small rectangles and triangles in the interior of ${\cal R}$, we conclude that Eq. (6) holds for arbitrary parallelograms.
