# Transpose of matrix

1. Jul 19, 2006

### Ara macao

$$(ABC)^T, A,B,C$$ are all symmetric, then why isn't $$(ABC)^T = CBA$$? If you consider that $$(ABC)^T = (C^T)(B^T)(A^T)$$ and in symmetrix cases, then $$C^T = C$$ and so on...?

(Latex edit by HallsofIvy)

Last edited by a moderator: Jul 28, 2006
2. Jul 19, 2006

### matt grime

Who says that (ABC)^T is not CBA when all three are symmetric?

3. Jul 19, 2006

### Ara macao

The solutions manual to Gilbert Strang Linear Algebra...

4. Jul 19, 2006

### matt grime

Right, why don't you post the full question and the full answer from this book? I mean, is the question:

Q. if A,B, and C are symmetric does (ABC)^T = CBA?
A. No.

5. Jul 19, 2006

### matt grime

Right, why don't you post the full question and the full answer from this book? I mean, is the question:

Q. if A,B, and C are symmetric does (ABC)^T = CBA?
A. No.

6. Jul 19, 2006

### Ara macao

Yes, that is the case

7. Jul 19, 2006

### matt grime

Then the asnwer book is wrong, if that is the precise statement of the question.

8. Jul 27, 2006

### Bob

ABC $$\neq$$CBA

Last edited: Jul 28, 2006
9. Jul 28, 2006

### matt grime

And no one is claiming that they are equal.