1. Sep 27, 2009

### yaganon

If AB = AC, it's possible that B /= C

conversely:
if BA = CA, is it possible that C /= B?

/= means doesn't equal

2. Sep 27, 2009

### nicksauce

Certainly. In both cases let A = 0, B = Anything, C = Anything else.

3. Sep 27, 2009

### HallsofIvy

Staff Emeritus
More generally, if A is a matrix that does not have an inverse, then we can have A(B- C)= 0 without B- C being equal to 0. And, similarly, (B- C)A= 0 without B- C= 0.

4. Sep 27, 2009

### yaganon

hallsofivy

why does A have to be singular?

5. Sep 28, 2009

### lurflurf

say
AB=AC
suppose A has an inverse D such that AD=DA=
DAB=DAC
B=C
thus if B/=C A must be singular

Also the matrices might not be square
[[1,1,1]]*[[1,1,1]]=[[1,1,1]]*[[3,0,0]]=[[3]]
[[1,1,1]]/=[[3,0,0]]

6. Sep 28, 2009

### trambolin

A basic example for you...

$\begin{pmatrix}0 & 0 & 0\\0 &1 & 0\\ 0 & 0 &0\end{pmatrix} = \begin{pmatrix}1 & 0 & 1\\0 &1 & 0\\ 1 & 0 &1\end{pmatrix} \begin{pmatrix}1 & 0 & -1\\0 &1 & 0\\ -1 & 0 &1\end{pmatrix} = \begin{pmatrix}1 & 0 & 1\\0 &1 & 0\\ 1 & 0 &1\end{pmatrix} \begin{pmatrix}2 & 0 & -5\\0 &1 & 0\\ -2 & 0 &5\end{pmatrix} = \begin{pmatrix}0 & 0 & 0\\0 &1 & 0\\ 0 & 0 &0\end{pmatrix}$

But

$\begin{pmatrix}1 & 0 & -1\\0 &1 & 0\\ -1 & 0 &1\end{pmatrix} \neq \begin{pmatrix}2 & 0 & -5\\0 &1 & 0\\ -2 & 0 &5\end{pmatrix}$

If you look at the rank of A and the vector that is in the nullspace of A, you can see why this holds with the columns of the B and C are chosen as such.

Last edited: Sep 28, 2009