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So, Ab(sub)n = 0, but b(sub)n != 0. I have the answer, but I want to make sure I understand it. Going back to Linear Independence section I see that an indexed set of vectors {v1, ... vp} in Rn is linearly dependent if there exist weights c1..., cp, not all zero such that c1v1 + .. + cpvp = 0

So, I figure, plug in 'bn' for 'cp' and make the set {v1, ... vp} be the columns of A, and you would have A is Linearly dependent if B has columns, not all zero, that make a1v1 + ... anvn = 0.

So the proof then is that since we know that B has columns not all zero (in fact, it has no zero columns at all), then A must be linearly dependent.

Is this the right way to figure this? They give the right answer, but not the logic in the answer key. Furthermore, is there a way to remember this type of thing? Is there some logical way to prove this for myself aside from just knowing the theorum?

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another question I don't get :

Let A = [2,5;-3,1] and B = [4,-5;3,k]. What value of k, if any, will make AB = BA.

I got as far as figuring that 6-3k = -9 and 5k-10 = 15. So, solving for each, I get k = 5 and k = -1. But, the answer is k-5. Is the wording of the question wrong, or what is my problem here? Is there a simple algebra trick to this?

Thanks!

Angela.