Symmetric Matrix Transpose: ABC^T ≠ CBA?

[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)

 

[matt grime]

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

 

[Ara macao]

The solutions manual to Gilbert Strang Linear Algebra...

 

[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.

 

[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.

 

[Ara macao]

Yes, that is the case

 

[matt grime]

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

 

[Bob]

ABC $$\neq$$CBA

 

[matt grime]

And no one is claiming that they are equal.