Prove or disprove that AB=AC $\implies$ B=C

Question

I proved it as follows but I'm not so sure about it.

A, B and C are square matrices of the same order.

Assume $ B \neq C $

$$ AB \neq AC$$ $$ B \neq C \implies AB \neq AC$$ $$ \neg ( AB \neq AC) \implies \neg ( B \neq C ) $$ $$AB =AC \implies B=C $$

In general, one cannot expect to prove anything about something concrete (here maybe the integers, or the reals, it has not been made clear) by manipulation of logical symbols alone. — André Nicolas, Aug 09 '14 at 06:49
What is the domain we're talking about? What does the $;.;$ operator mean? — MarnixKlooster ReinstateMonica, Aug 09 '14 at 07:07
The claim that $B\ne C \implies A.B \ne A.C$ is probably incorrect. Although as @MarnixKlooster says, it's impossible to say without defining what kind of objects $A$, $B$, and $C$ are and what the $.$ operator is. — augurar, Aug 09 '14 at 07:37

score 3 · Answer 1 · answered Aug 09 '14 at 06:08

3

What if $A$ is zero? ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

answered Aug 09 '14 at 06:08

anon

2

What if $.$ is addition? You shouldn't post an answer to an underspecified question. – augurar Aug 09 '14 at 07:40
4

With the balance of probability beyond a given threshold I think it makes sense to assume the most obvious choice. This one is perhaps debatable; if OP gives any indication to the contrary I'd happily delete. (For my side of the debate: the title doesn't even have a period symbol, and usually concatenation implies a multiplication operation, and usually a multiplication operation is in the context of numbers, especially when no other context is given.) – anon Aug 09 '14 at 07:43
You're probably right, but one immediate counterexample is if $A$, $B$, and $C$ are members of a group, in which case the claim is true. – augurar Aug 09 '14 at 07:50
"." Is not addition, I never thought one would consider it as a different operator. – S.Dan Aug 09 '14 at 08:47

score -2 · Answer 2 · answered Aug 09 '14 at 06:32

-2

If A be non-zero,then it is true.

answered Aug 09 '14 at 06:32

Mehdi Sajjadian

1

Depends on the context. There are many valid contexts, e.g. in abstract algebra, where your claim is simply wrong. The main problem is that the OP did not specify what kind of objects $A,B,C $ are, and what $AB$ means et cetera. – Jyrki Lahtonen Aug 09 '14 at 07:36
2

What if $A$ is nonzero and singular? – Algebraic Pavel Aug 09 '14 at 10:47

2 Answers2