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 \neqCBA

 

[matt grime]

And no one is claiming that they are equal.