The discussion focuses on analyzing a linear transformation represented by matrix \( D \) in the context of a parallelogram defined by vertices \( O(0,0) \), \( A(1,0) \), \( B(1,1) \), and \( C(0,1) \). The correct columns of matrix \( D \) are determined by mapping \( \overrightarrow{OA} \) to \( \overrightarrow{OA'} \) and \( \overrightarrow{OC} \) to \( \overrightarrow{OC'} \), resulting in \( D = \begin{pmatrix}2&6\\2&4\end{pmatrix} \). The discussion emphasizes that incorrect choices of columns, such as including vertex \( B'(8,6) \), lead to erroneous mappings, specifically yielding an incorrect output for \( \overrightarrow{OB} \).
PREREQUISITESStudents and educators in mathematics, particularly those studying linear algebra, as well as professionals working with linear transformations in computer graphics or data analysis.
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
