Solving Matrix Multiplication: A*B^T with A=2,1;5,3 and B=1,-1,1;-1,1,1

AI Thread Summary
The discussion centers on the multiplication of matrix A by the transpose of matrix B, specifically A*B^T, where A is a 2x2 matrix and B is a 2x3 matrix. The confusion arises from the dimensions of the matrices, as the transpose of B results in a 3x2 matrix, making it impossible to multiply with A. Participants clarify that the operation cannot be performed as stated, and suggest that the intended operation might have been (AB)^T instead. The realization that the order of multiplication matters is acknowledged, leading to a better understanding of matrix operations. Ultimately, the conclusion is that A*B^T cannot be computed due to incompatible dimensions.
lagwagon555
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Homework Statement


Find A*B^T, where A=

2 1
5 3

And B=

1 -1 1
-1 1 1


Homework Equations



n/a



The Attempt at a Solution



The problem is, is that if I get the transpose of B, I will end up with a 3x2 matrix. Isn't this impossible to multiply by a 2x2 matrix? Maybe the answers is just 'not possible', but I have a feeling I'm doing something wrong. Anybody got any pointers? Thanks!
 
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You are right, A*BT can't be computed. (AB)T can though, are you sure you weren't supposed to calculate that?
 
The way it was actually written was BTA, so I think it made a point of showing that it wasn't (AB)T. So I'm thinking that the answer is that it can't be computed? Thanks for the quick reply!

EDIT: ooooooooooh. Ooops. How embarassing. Of course it makes a difference when B comes first. Argh!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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