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: 16

Answer

Geometrically, the solution set of the homogenous equation is translated by vector $\begin{bmatrix} -5\\ 3\\ 0 \end{bmatrix}$

Work Step by Step

$\begin{bmatrix} 1 & 3 & -5 & 4\\ 1 & 4 & -8 & 7\\ -3 & -7 & 9 & -6 \end{bmatrix} $ Subtract row 1 from row 2 $\begin{bmatrix} 1 & 3 & -5 & 4\\ 0 & 1 & -3 & 3\\ -3 & -7 & 9 & -6 \end{bmatrix} $ Add 3 times row 1 to row 3 $\begin{bmatrix} 1 & 3 & -5 & 4\\ 0 & 1 & -3 & 3\\ 0 & 2 & -6 & 6 \end{bmatrix} $ Subtract 2 times row 2 from row 3 $\begin{bmatrix} 1 & 3 & -5 & 4\\ 0 & 1 & -3 & 3\\ 0 & 0 & 0 & 0 \end{bmatrix} $ Subtract 3 times row 2 from row 1 $\begin{bmatrix} 1 & 0 & 4 & -5\\ 0 & 1 & -3 & 3\\ 0 & 0 & 0 & 0 \end{bmatrix} $ Subtract 3 times row 2 from row 1 $x_1=-5-4x_3$ $x_2=3+3x_3$ $x_3$ is free $x=\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} -5-4x_3\\ 3+3x_3\\ x_3 \end{bmatrix}=\begin{bmatrix} -5\\ 3\\ 0 \end{bmatrix}+x_3\begin{bmatrix} -4\\ 3\\ 1 \end{bmatrix} $ This means geometrically, the solution set of the homogenous equation is translated by vector $\begin{bmatrix} -5\\ 3\\ 0 \end{bmatrix}$
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