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 - 1.4 Exercises - Page 41: 17

Answer

No, by theorem 4.(d and a).

Work Step by Step

Row reducing A, $\left[\begin{array}{llll} 1 & 3 & 0 & 3\\ -1 & -1 & -1 & 1\\ 0 & -4 & 2 & -8\\ 2 & 0 & 3 & 1 \end{array}\right]\left(\begin{array}{l} .\\ +\mathrm{r}_{1}.\\ .\\ -2\mathrm{r}_{1}. \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 3 & 0 & 3\\ 0 & 2 & -1 & 4\\ 0 & -4 & 2 & -8\\ 0 & -6 & 3 & -7 \end{array}\right]\left(\begin{array}{l} .\\ .\\ +2\mathrm{r}_{2}.\\ +3\mathrm{r}_{2}. \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 3 & 0 & 3\\ 0 & 2 & -1 & 4\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 5 \end{array}\right]\left(\begin{array}{l} .\\ .\\ \leftrightarrow \mathrm{r}_{4}.\\ . \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 3 & 0 & 3\\ 0 & 2 & -1 & 4\\ 0 & 0 & 0 & 5\\ 0 & 0 & 0 & 0 \end{array}\right]\qquad $3 pivot positions... Row 4 does not have a pivot position. In Theorem 4, $\mathrm{d}. A$ has a pivot position in every row, is false. So then, is $\mathrm{a}$. For each $\mathrm{b}$ in $\mathbb{R}^{m}$, the equation $A\mathrm{x}=\mathrm{b}$ has a solution (false)
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