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 - 1.8 Exercises - Page 69: 9

Answer

$x=x_3\begin{bmatrix} 9\\ 4\\ 1\\ 0 \end{bmatrix}+x_4\begin{bmatrix} -7\\ -3\\ 0\\ 1 \end{bmatrix}$

Work Step by Step

$A=\begin{bmatrix} 1 & -4 & 7 & -5\\ 0 & 1 & -4 & 3\\ 2 & -6 & 6 & -4 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix}$ $\begin{bmatrix} 1 & -4 & 7 & -5 & 0\\ 0 & 1 & -4 & 3 & 0\\ 2 & -6 & 6 & -4 & 0 \end{bmatrix}$~$\begin{bmatrix} 1 & 0 & -9 & 7 & 0\\ 0 & 1 & -4 & 3 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$ $x_1=9x_3-7x_4$ $x_2=4x_3-3x_4$ $x_3$ is free $x_4$ is free $x=x_3\begin{bmatrix} 9\\ 4\\ 1\\ 0 \end{bmatrix}+x_4\begin{bmatrix} -7\\ -3\\ 0\\ 1 \end{bmatrix}$
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