The columns of $D$ can be determined by using any two of the other vertices of the parallelogram.
It would be wrong to choose $(8,6)$ as a column, so not all of the remaining three vertices can serve as columns. Let $O$ be the origin, $A(1,0)$, $B(1,1)$ and $C(0,1)$. Let also $A'(2,2)$, $B'(8,6)$ and $C'(6,4)$. Then $D$ should map $\overrightarrow{OA}$ to $\overrightarrow{OA'}$ and $\overrightarrow{OC}$ to $\overrightarrow{OC'}$. Then $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}$ will automatically be mapped to $\overrightarrow{OA'}+\overrightarrow{OC'}=\overrightarrow{OB'}$. Alternatively, $\overrightarrow{OA}$ can be mapped to $\overrightarrow{OC'}$ and $\overrightarrow{OC}$ to $\overrightarrow{OA'}$; the location of $B'$ will be the same. In the first scenario
\[
D\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}2\\2\end{pmatrix},
\]
so the first column of $D$ is $(2,2)$. Also
\[
D\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}6\\4\end{pmatrix},
\]
so the second column of $D$ is $(6,4)$. In the second scenario the columns are swapped.
But if, say, $D=\begin{pmatrix}2&8\\2&6\end{pmatrix}$, then $D$ maps $\overrightarrow{OA}$ to $\overrightarrow{OA'}$ and $\overrightarrow{OC}$ to $\overrightarrow{OB'}$. In this case, $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}$ will be mapped to $\overrightarrow{OA'}+\overrightarrow{OB'}$, and the coordinates of the resulting vector, i.e., the coordinates of the image of $B$, are $(10,8)$, which is incorrect.
