Does anyone have any advice on how to solve harder(not necessarily in the knowledge of mathematics required- I've only covered a course in Algebra which had no group theory in it!) Linear Algebra/Algebra problems? I always find Algebra a fair bit more challenging than Analysis and I need all the help I can get with it!
As an example of what I mean, consider the following question:
Given square matrices A,B,C with A necessarily invertible, prove that if (A-B)C=BA^-1 then C(A-B)=A^-1 B.
I spent 4 and a half hours on this particular question and still didn't come up with the trick for solving it. Is there any particular way of getting better at these type of problems or is it just a lot of practice?
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- Thread Starter
- 08-08-2009 23:53
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- Wiki Support Team
- 09-08-2009 01:03
Right. Well, this has taken me about an hour to solve. Solution in spoiler below in steps. And, to be honest, I can't tell you anything about how I solved it; I just tried lots of things until it came out. I noticed that, when I expanded out the brackets of (A - B)C = BA^-1, I got something that looked kind-of-almost-factorisable, and products are easier to deal with than sums. So eventually I tried stuff and got there. I don't think there's anything easier than that.
Spoiler:ShowTry and find a product that looks something like (A - B)C = BA^-1.
If you can't find it, here it is:
Spoiler:ShowConsider (A - B)(A^-1 + C), which I've just pulled out of thin air.
Rest of solution:
Spoiler:ShowThis is equal to I - BA^-1 - BC + AC. But if (A-B)C = BA^-1, then -BA^-1 - BC + AC = 0 (expand the brackets), and so (A - B)(A^-1 + C) = I. By a simple argument about rank/nullity, we must have that (A - B) and (A^-1 + C) have full rank, and so have inverses, and so are inverses of each other. So clearly (A^-1 + C)(A - B) = I, and expand that.
- 10-08-2009 13:41
please note that this was a problem set in this years IMC.
something ******** OP decided not to post. nice work though billeh.