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: 27

Answer

$u^Tv=\begin{bmatrix} -2a+3b-4c \end{bmatrix} $ $v^Tu=\begin{bmatrix} -2a+3b-4c \end{bmatrix} $ $uv^T=\begin{bmatrix} -2a&-2b&-2c\\ 3a&3b&3c\\ -4a&-4b&-4c \end{bmatrix} $ $vu^T=\begin{bmatrix} -2a&3a&-4a\\ -2b&3b&-4b\\ -2c&3c&-4c \end{bmatrix} $

Work Step by Step

$u=\begin{bmatrix} -2\\ 3\\ -4 \end{bmatrix} $$v=\begin{bmatrix} a\\ b\\ c \end{bmatrix} $$u^T=\begin{bmatrix} -2&3&-4 \end{bmatrix} $$v^T=\begin{bmatrix} a&b&c \end{bmatrix} $ $u^Tv=\begin{bmatrix} -2a+3b-4c \end{bmatrix} $ $v^Tu=\begin{bmatrix} -2a+3b-4c \end{bmatrix} $ $uv^T=\begin{bmatrix} -2a&-2b&-2c\\ 3a&3b&3c\\ -4a&-4b&-4c \end{bmatrix} $ $vu^T=\begin{bmatrix} -2a&3a&-4a\\ -2b&3b&-4b\\ -2c&3c&-4c \end{bmatrix} $
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