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anonymous
 5 years ago
Prove that whenever A and B are matrices for which AB is defined, then (B^T)(A^T) is also defined. Then show (AB)^T=(B^T)(A^T). (transpose)
anonymous
 5 years ago
Prove that whenever A and B are matrices for which AB is defined, then (B^T)(A^T) is also defined. Then show (AB)^T=(B^T)(A^T). (transpose)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So what does it mean AB is defined?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm guessing, if A is an m x n matrix and B is an n x p matrix

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...then AB will be defined as the m x p matrix

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well my book says the transpose of a product of any number of matrices is the product of the tranposes in the reverse order.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my prof always starts of a proof asking "What do we know"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So we know that AB is defined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess i'm having trouble understanding this whole "ijth" entry thing, I didn't understand it much in lecture and my book doesn't expand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you tell me what topic it is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nDimensional Geometry: Matrix Multiplication the book says that the matrix product AB is the mxp matrix whose ijth entry is the dot product of the ith row of A and the jth column of B (considered as vectors in R^n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0check out this webpage it explains matrix multiplication in easy to understand language http://www.purplemath.com/modules/mtrxmult.htm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a little, i looked into my book a bit more and convinced myself that the ijth entry of AB is \[\sum_{k}^{n}\] \[a _{ik}b _{kj}\] , which equals the jith entry of \[B ^{T}A ^{T}\] which is \[\sum_{k}^{n} b _{kj}a _{ik}\]
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