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## Homework Statement

Determine which of the formulas hold for all invertible nxn matrices A and B:

(a) (A+A

^{-1})

^{8}=A

^{8}+A

^{-8}

(b) A + I

_{n}in invertible

(c) AB=BA

(d) (A+B)(A-B) = A

^{2}- B

^{2}

(e) (ABA

^{-1})

^{3}= AB

^{3}A

^{-1}

(f) A

^{6}B

^{4}in invertible

## The Attempt at a Solution

(a) I don't know this one, I don't think it's true, either. Because if you just square the thing, you'll get:

A

^{2}+ 2I + (A

^{-1})

^{2}, and if you continue this way, you'll just get more and more of Is in there.

(b)A + I

_{n}is invertible. I'm pretty sure with that one. Because I is a square matrix, same size as A (both are nxn) and if you add you will still have nxn, meaning it will be invertible.

(c)I don't think AB=BA. I tried it out with two sample matrices (just random matrices) and it doesn't work.

(d)Since AB=BA isn't true, then this should also be false. Because when you expand you will have -AB+BA and unless they equal to each other, these terms don't cancel out.

(e)I think this is true, because A*A

^{-1}is I, so we essentially have (IB)

^{3}, which equals to IB

^{3}, and if we were to put A*A

^{-1}back in, we'd see that the two sides do indeed equal.

(f)Yep. A and B are the same size and square. So no matter how many times we raise each one of them to whatever power, they will still remain the same size and will be invertible.

However, that least one of my "logical" answers is wrong. Can you point out what assumption I'm making that's wrong? Thanks!