Answer
The linear mapping is $G = \left( {4u + 2v,u + 3v} \right)$.
Work Step by Step
A linear mapping is given by the form:
$G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$
The domain ${\cal R}:\left[ {0,1} \right] \times \left[ {0,1} \right]$ in the $uv$-plane is spanned by the vectors $\overrightarrow {OP} = \left( {0,1} \right)$ and $\overrightarrow {OQ} = \left( {1,0} \right)$. Using $G\left( {u,v} \right)$ we obtain the images of $\overrightarrow {OP} $ and $\overrightarrow {OQ} $:
$G\left( {\overrightarrow {OP} } \right) = G\left( {0,1} \right) = \left( {C,D} \right)$
$G\left( {\overrightarrow {OQ} } \right) = G\left( {1,0} \right) = \left( {A,B} \right)$
Since $G$ is a linear map, $G\left( {\overrightarrow {OP} } \right)$ and $G\left( {\overrightarrow {OQ} } \right)$ span the parallelogram in the $xy$-plane. Thus,
$G\left( {\overrightarrow {OP} } \right) = \left( {C,D} \right) = \left( {2,3} \right)$
$G\left( {\overrightarrow {OQ} } \right) = \left( {A,B} \right) = \left( {4,1} \right)$
Therefore, $A=4$, $B=1$, $C=2$, and $D=3$.
Thus, we obtain the linear mapping: $G = \left( {4u + 2v,u + 3v} \right)$.
