Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises - Page 62: 31

Answer

$ \begin{bmatrix}1\\1\\-1\end{bmatrix} $

Work Step by Step

Since the third column of the matrix is the sum of the first two, we have the following: $ \begin{bmatrix}2&3&5\\-5&1&-4\\-3&-1&-4\\1&0&1\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix}=1\cdot\begin{bmatrix}2\\-5\\-3\\1\end{bmatrix}+1\cdot\begin{bmatrix}3\\1\\-1\\0\end{bmatrix}+(-1)\cdot\begin{bmatrix}5\\-4\\-4\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix} $. In fact, any vector of the form $(c,c,-c)$, $c\neq 0$, is a nontrivial solution to the given homogeneous equation.
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