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: 7

Answer

$\mathbf{x}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}=x_{3}\begin{bmatrix}-9\\4\\1\\0\end{bmatrix}+x_{4}\begin{bmatrix}8\\-5\\0\\1\end{bmatrix}$

Work Step by Step

To reduce the matrix to rref, we simply add $-3$ times the second row to the first, yielding $\begin{bmatrix}1&0&9&-8\\0&1&-4&5\end{bmatrix}$. This is equivalent to the set of equations $\begin{cases}x_{1}+9x_{3}-8x_{4}=0\\x_{2}-4x_{3}+5x_{4}=0\end{cases}$. We solve for $x_{1}$ and $x_{2}$ in terms of $x_{3}$ and $x_{4}$, $\begin{cases}x_{1}=-9x_{3}+8x_{4}\\x_{2}=4x_{3}-5x_{4}\\x_{3}=x_{3}\\x_{4}=x_{4}\end{cases}$, which we finally rewrite as a single vector equation.
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