2007-12-28 17:43:36 · 6 answers · asked by UnknownD 6 in Science & Mathematics ➔ Mathematics
*cough cough*
Let your wild imagination come to life.
2007-12-28 17:52:32 · update #1
Dang, you're boring people. Nobody said something stupid T-T.
2007-12-28 17:54:28 · update #2
How about when A is a 1x2 matrix and B is a 2x1 matrix?
A*B would result in a 1x1 matrix, but B*A would give you a 2x2 matrix?
2007-12-28 17:52:16 · answer #1 · answered by buaya123 3 · 4⤊ 1⤋
Suppose AB is an element of the free group on two generators (which has presentation < A , B | >).
Then AB does not equal BA. If it did, we'd have:
AB=BA
implies
ABA^(-1)B^(-1) = 1
But this is not possible since the presentation contains no relations. Therefore, AB doesn't equal BA.
2007-12-28 18:07:21 · answer #2 · answered by jtabbsvt 5 · 0⤊ 0⤋
Vector cross products is another example, in that case,
A × B = -B × A
and if you go in for group theory, you can find tons of non-Abelian groups where A * B ≠ B * A
2007-12-28 17:59:30 · answer #3 · answered by devilsadvocate1728 6 · 2⤊ 0⤋
If either A or B is the square root of any number > 0, it could be the positive root on one side and the negative root on the other side of the inequality.
2007-12-28 18:31:46 · answer #4 · answered by Arnold K 2 · 0⤊ 1⤋
asking for a counterexample to an algebraic property is silly.
In the set of complex numbers there are no such ways.
So, what are A and B elements of? A group? a ring?
Are they matrices?
Not enough information to answer but as far as common usage goes, impossible.
2007-12-28 17:51:24 · answer #5 · answered by kozzm0 7 · 0⤊ 2⤋
Multiplication is really about groups.
One group of ten beautiful woman is definitely not equal to ten groups of one beautiful woman.
2007-12-28 19:59:23 · answer #6 · answered by McMurphyRP 3 · 0⤊ 0⤋