## Omar91X 2 years ago Find nonzero matrices A, B, and C such that AC = BC and A does not equal B

1. hewsmike

$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.

2. Omar91X

Thank you sir, I appreciate the explanation.

3. seiga

^he's right.