Here's the question you clicked on:
Omar91X
Find nonzero matrices A, B, and C such that AC = BC and A does not equal B
\[A = \left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\]\[B = \left[\begin{matrix}1 & 2 \\ 3 & 6\end{matrix}\right]\]\[C = \left[\begin{matrix}1 & 0 \\ 0 & 0\end{matrix}\right]\]so \[AC = \left[\begin{matrix}1 & 2 \\ 3 & 4\end{matrix}\right]\left[\begin{matrix}1 &0 \\ 0 & 0\end{matrix}\right]=\left[\begin{matrix}1 &0 \\3 & 0\end{matrix}\right]\]and\[BC = \left[\begin{matrix}1 & 2 \\ 3 & 6\end{matrix}\right]\left[\begin{matrix}1 &0 \\ 0 & 0\end{matrix}\right]=\left[\begin{matrix}1 &0 \\3 & 0\end{matrix}\right]\]there are many choices here. The key issue is that C must be singular ( non-invertible ) because if it was then \[A C = B C\]becomes\[A C C^{-1}= B C C^{-1}\]\[A I = B I\]\[A = B\]which the question does not allow. The values in the second columns of A and/or B are irrelevant, as C is selecting only their first columns and hence the products are equal.
Thank you sir, I appreciate the explanation.