1. Let $ A \in M_{m \times p} $ and $ B \in M_{p \times n} $. Recall that the definition of matrix multiplication gives that the $(i, j)$-entry of the product $ AB $ is\[[AB]_{ij} = \sum_{k=1}^{p} A_{ik}B_{kj}.\]Use the definition of matrix multiplication to show that $ (AB)^T = B^T A^T $.

Given matrices are Now from property of product of matrices ,Now we have to show that .Now this is…

Answered: 1. Let A € Mmxp and B € Mpxn. Recall…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

1. Let $ A \in M_{m \times p} $ and $ B \in M_{p \times n} $. Recall that the definition of matrix multiplication gives that the $(i, j)$-entry of the product $ AB $ is

\[
[AB]_{ij} = \sum_{k=1}^{p} A_{ik}B_{kj}.
\]

Use the definition of matrix multiplication to show that $ (AB)^T = B^T A^T $.

Expert Solution

Step 1: matrix multiplication related problem

Given matrices are $A element of M subscript m cross times p end subscript space a n d space B element of M subscript p cross times n end subscript$

Now from property of product of matrices , $open square brackets A B close square brackets subscript i j end subscript equals sum from k equals 1 to p of A subscript i k end subscript B subscript k j end subscript$

Now we have to show that $open parentheses A B close parentheses to the power of T equals B to the power of T A to the power of T$ .

Now $open square brackets open parentheses A B close parentheses to the power of T close square brackets subscript i j end subscript equals open square brackets A B close square brackets subscript j i end subscript equals sum from k equals 1 to p of A subscript j k end subscript B subscript k i end subscript equals sum from k equals 1 to p of B subscript k i end subscript A subscript j k end subscript equals sum from k equals 1 to p of B to the power of T subscript i k end subscript A to the power of T subscript k j end subscript equals open square brackets B to the power of T A to the power of T close square brackets subscript i j comma end subscript$ this is true for all $1 less or equal than i less or equal than n space a n d space 1 less or equal than j less or equal than m$

Thus we get $i j space t h space e l e m e n t space o f space open parentheses A B close parentheses to the power of T equals i j space t h space e l e m e n t space o f space B to the power of T A to the power of T$