Symmetric Matrix Transpose: ABC^T ≠ CBA?

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Homework Help Overview

The discussion revolves around the properties of symmetric matrices and the transpose operation, specifically questioning the relationship between the transpose of the product of three symmetric matrices and the product of their transposes in a different order.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore whether the equality (ABC)^T = CBA holds when A, B, and C are symmetric matrices, questioning the validity of a claim from a solutions manual.

Discussion Status

There is an active exploration of the question, with participants expressing differing views on the correctness of the solutions manual's assertion. Some participants are seeking clarification on the original question posed in the manual.

Contextual Notes

Participants note the importance of the precise wording of the question from the solutions manual, which is central to the discussion. There is also mention of a potential misunderstanding regarding the equality of the two expressions.

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

(Latex edit by HallsofIvy)
 
Last edited by a moderator:
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Who says that (ABC)^T is not CBA when all three are symmetric?
 
The solutions manual to Gilbert Strang Linear Algebra...
 
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.
 
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.
 
Yes, that is the case
 
Then the asnwer book is wrong, if that is the precise statement of the question.
 
ABC [tex]\neq[/tex]CBA
 
Last edited:
And no one is claiming that they are equal.
 

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