Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 2 - Matrix Algebra - 2.1 Exercises - Page 103: 33

Answer

The $(i,j)$-entry of $(AB)^{T}$ is the $(j, i)$-entry of $AB$, which is the inner product of (row j in A) and (column i in B) $[(AB)^{T}]_{ij}=a_{j1}b_{1i}+\cdots+a_{jn}b_{ni}$ On the other hand, the $(i,j)$-entry of $B^{T}A^{T}$ equals the inner product of (row i in $B^{T}$) and (column j in $A^{T}$), equals the inner product of (column i in $B$) and (row j in $A$), equals the inner product of (row j in $A$) and (column i in $B$) or, $a_{j1}b_{1i}+\cdots+a_{jn}b_{ni}$ This proves that $(AB)^{T}=B^{T}A^{T}$

Work Step by Step

The answer contains the proof of: $(AB)^{T}=B^{T}A^{T}$
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