Answer
The translate of the linear mapping:
$G\left( {u,v} \right) = \left( {4 + 3u - 4v,2 + 9u + 6v} \right)$
Work Step by Step
Step 1. Find the linear mapping ${G_0}$ that maps ${{\cal R}_0} = \left[ {0,1} \right] \times \left[ {0,1} \right]$ to the parallelogram spanned by the vectors $\left( {3,9} \right)$ and $\left( { - 4,6} \right)$ based at the origin.
Let the parallelogram in the $xy$-plane be ${{\cal D}_0}$.
The linear map is given by the form:
${G_0}\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
The domain ${{\cal R}_0}:\left[ {0,1} \right] \times \left[ {0,1} \right]$ in the $uv$-plane is spanned by the vectors $\overrightarrow {OP} = \left( {1,0} \right)$ and $\overrightarrow {OQ} = \left( {0,1} \right)$. Using ${G_0}\left( {u,v} \right)$ we obtain the images of $\overrightarrow {OP} $ and $\overrightarrow {OQ} $:
${G_0}\left( {\overrightarrow {OP} } \right) = {G_0}\left( {1,0} \right) = \left( {A,B} \right)$
${G_0}\left( {\overrightarrow {OQ} } \right) = {G_0}\left( {0,1} \right) = \left( {C,D} \right)$
Since ${G_0}$ is a linear map, ${G_0}\left( {\overrightarrow {OP} } \right)$ and ${G_0}\left( {\overrightarrow {OQ} } \right)$ span the parallelogram in the $xy$-plane. Thus,
${G_0}\left( {\overrightarrow {OP} } \right) = \left( {A,B} \right) = \left( {3,9} \right)$
${G_0}\left( {\overrightarrow {OQ} } \right) = \left( {C,D} \right) = \left( { - 4,6} \right)$
From these equations, we obtain: $A=3$, $B=9$, $C=-4$, and $D=6$.
Thus, we obtain the linear map:
${G_0} = \left( {3u - 4v,9u + 6v} \right)$
Step 2. Translate the linear map ${G_0}$ such that ${{\cal D}_0}$ is translated to ${\cal D}$ whose base is at $\left( {4,2} \right)$.
In this case $a=4$ and $b=2$. Thus, the translate of ${G_0}$ is $G$ given by
$G\left( {u,v} \right) = \left( {a + Au + Cv,b + Bu + Dv} \right)$
$G\left( {u,v} \right) = \left( {4 + 3u - 4v,2 + 9u + 6v} \right)$
