Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.5 Exercises - Page 48: 13

Answer

Geometrically, the solution set is the line through$\left[\begin{array}{ r }5\\-2\\0\end{array}\right]\text{ in the direction of }\left[\begin{array}{ r }4\\-7\\1\end{array}\right].$

Work Step by Step

To write the general solution in parametric vector form, pull out the constant terms that do not involve the free variable: $x=\left[\begin{array}{ l }x_1\\x_2\\x_3\end{array}\right]=\left[\begin{array}{ c }5+4x_3\\-2-7x_3\\x_3\end{array}\right]=\left[\begin{array}{ r }5\\-2\\0\end{array}\right]+\left[\begin{array}{ r }4x_3\\-7x_3\\x_3\end{array}\right]=\left[\begin{array}{ r }5\\-2\\0\end{array}\right]+x_3\left[\begin{array}{ r }4\\-7\\1\end{array}\right]$ $=p+x_3q$ Geometrically, the solution set is the line through$\left[\begin{array}{ r }5\\-2\\0\end{array}\right]\text{ in the direction of }\left[\begin{array}{ r }4\\-7\\1\end{array}\right].$
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