# Linear Algebra-Inverses

1. Feb 27, 2007

### craigy

1. The problem statement, all variables and given/known data

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

1. (In-A)(In+A) = In - A^2
2. (A+B)(A-B) = A^2 - B^2
3. A^8*B^5 is invertible
4. (A+A^-1)^9 = A^9 + A^-9
5. AB = BA
6. A+A^-1 is invertible

I know 5 is right, and number 2 wrong, but what else?

2. Feb 27, 2007

A starting hint: a product of regular matrices is a regular matrix.

3. Feb 27, 2007

### craigy

I know that but that doesn't solve the problem

4. Feb 28, 2007

### HallsofIvy

For 1, 4, and 6, just go ahead and do the multiplication! What is (I-A)(I+A)?
If you don't want to actually multiply (A+A-1) nine times, try one or two times and see if you can't find a pattern. That's how you solve math problems- you try things, you don't just sit and try to "remember" how to solve it! As for (A+ A-1) try to guess an inverse and then do the calculation to see what happens. Try it with a simple diagonal matrix first. (Another general math method- to do complex problems, try a few simple examples first.)

5. Feb 28, 2007

### craigy

I sub n is an identity matrix. 2, 3 and 5 seems to work when using a diagonal matrix.

6. Feb 28, 2007

You're not interested in diagonal matrices, you're interested in all regular (i.e. invertible) matrices. Take, for example $$A=\left(\begin{array}{cc}1 & 2\\0 & -1\end{array}\right)$$, and $$B=\left(\begin{array}{cc}-1 & 0\\2 & 3\end{array}\right)$$. Obviously $$AB\neq BA$$, although both det(A) and det(B) are non-zero.

Edit: after reading this, you should easily see if (2) holds or not for two regular nxn matrices A and B.

Last edited: Feb 28, 2007
7. Feb 28, 2007

### craigy

Thanks radou, after some thought, I figured out that 1 and 3 are the only formulas that hold for all invertible n x n matrices A and B.

8. Mar 1, 2007

Assuming that by " (In-A)(In+A) = In - A^2 " you meant " (In-A)(In+A) = In^2 - A^2 ", where In = A^-1.

9. Mar 1, 2007

### HallsofIvy

In= A-1? Why would you assume that? Craigy specifically said In is the n by n identity matrix.
(In- A)(In+ A)= In2+ InA- AIn+ A2. Since InA= AIn= A, and In2= In, yes, (In- A)(In+ A)= In- A2.

10. Mar 1, 2007