Linear Algebra and Its Applications (5th Edition)

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

Chapter 6 - Orthogonality and Least Squares - 6.5 Exercises - Page 368: 13

Answer

Because the distance from Av to b is smaller than the distance from Au to b, u cannot be the least-squares solution to the equation.

Work Step by Step

$Au=\begin{bmatrix} 11\\-11\\11 \end{bmatrix}$ $||Au-b||=\sqrt{40}$ $Av=\begin{bmatrix} 7\\-12\\7 \end{bmatrix}$ $||Av-b||=\sqrt{29}$ Because the distance from Av to b is smaller than the distance from Au to b, u cannot be the least-squares solution to the equation.
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