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Matrix Multiplication Proof Q (1 Viewer)

VBN2470

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Write out entries for A and X and perform the matrix multiplication for AX and XA respectively. Then equate the matrices and you should get values for you entries in A. Your A matrix should then be a scalar multiple of I. Not sure if there is a faster way.
 

mreditor16

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Haha, yeh I tried that before you posted, but it went nowhere.....



Now what lol?
 

InteGrand

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Haha, yeh I tried that before you posted, but it went nowhere.....



Now what lol?
You've shown that cg = hb. Since this must be true for ALL real choices of g and h, c must equal b, and both must be 0.

Now use this fact in the other equalities you can derive. You should end up being able to show that a, b, c and d are all equal, which is what we're aiming to show.

Edit: I mean, we should be able to show that a = d and b = c = 0.
 
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InteGrand

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(If c and b were not both 0, it would be easy to choose a pair (g, h) that would make , a contradiction.)
 
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VBN2470

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You've shown that cg = hb. Since this must be true for ALL real choices of g and h, c must equal b.

Now use this fact in the other equalities you can derive. You should end up being able to show that a, b, c and d are all equal, which is what we're aiming to show.
You mean we're trying to show a and d are equal, whilst b = c = 0?
 
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InteGrand

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How does this work?



Why can't g=h=0?
Since we are told that AX = XA for any choice of X, we must have cg = hb for any choice of g and h. g and h are free for us to choose (as we can let X be anything) and the equality must always be maintained. In order for this equality to be maintained for all choices of g and h (hence including non-zero choices), we need c = b = 0.
 
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