Answer
Please see the figure attached.
The translate of the linear mapping is
$G\left( {u,v} \right) = \left( {1 + 2u + v,1 + 5u + 3v} \right)$
Work Step by Step
We sketch the parallelogram ${\cal P}$ in $xy$-plane with vertices $\left( {1,1} \right)$, $\left( {2,4} \right)$, $\left( {3,6} \right)$, $\left( {4,9} \right)$. From the figure attached we see that ${\cal P}$ is translated by $\left( {1,1} \right)$ from the origin. So, $a=1$, $b=1$.
The translate of a linear mapping has the form:
$G\left( {u,v} \right) = \left( {a + Au + Cv,b + Bu + Dv} \right)$
Substituting $a=1$ and $b=1$ in $G$ we get
$G\left( {u,v} \right) = \left( {1 + Au + Cv,1 + Bu + Dv} \right)$
The domain ${{\cal R}_0} = \left[ {0,1} \right] \times \left[ {0,1} \right]$ has vertices $\left( {0,0} \right)$, $\left( {0,1} \right)$, $\left( {1,1} \right)$, $\left( {1,0} \right)$.
Notice that ${\cal R}$ is spanned by $\left( {1,0} \right)$ and $\left( {0,1} \right)$ and ${\cal P}$ is spanned by $\left( {3,6} \right)$ and $\left( {2,4} \right)$.
Substituting these vertices in $G$ and equate each with the corresponding vertex in the $xy$-plane, we get
$G\left( {0,0} \right) = \left( {1,1} \right)$
$G\left( {0,1} \right) = \left( {1 + C,1 + D} \right) = \left( {2,4} \right)$
$G\left( {1,1} \right) = \left( {1 + A + C,1 + B + D} \right) = \left( {4,9} \right)$
$G\left( {1,0} \right) = \left( {1 + A,1 + B} \right) = \left( {3,6} \right)$
From these results, we obtain $A=2$, $B=5$, $C=1$, $D=3$.
Therefore, the translate of the linear mapping is
$G\left( {u,v} \right) = \left( {1 + 2u + v,1 + 5u + 3v} \right)$
