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: 20

Answer

Not every $\mathrm{b}\in \mathbb{R}^{4}$ can be written as a linear combination of the columns of B. No, the columns in B are vectors of $\mathbb{R}^{4}$, not $\mathbb{R}^{3}$and they span neither.

Work Step by Step

Row reducing B, $\left[\begin{array}{llll} 1 & 4 & 1 & 2\\ 0 & 1 & 3 & 4\\ 0 & 2 & 6 & 7\\ 2 & 9 & 5 & -7 \end{array}\right]\left(\begin{array}{l} .\\ .\\ .\\ -2\mathrm{r}_{1}. \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 4 & 1 & 2\\ 0 & 1 & 3 & 4\\ 0 & 2 & 6 & 7\\ 0 & 1 & 3 & -11 \end{array}\right]\left(\begin{array}{l} .\\ .\\ -2r_{2}.\\ -\mathrm{r}_{2}. \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 4 & 1 & 2\\ 0 & 1 & 3 & 4\\ 0 & 0 & 0 & 15\\ 0 & 0 & 0 & -7 \end{array}\right]\left(\begin{array}{l} .\\ .\\ .\\ +\frac{7}{15}\mathrm{r}_{3}. \end{array}\right)$ $\sim\left[\begin{array}{llll} 1 & 4 & 1 & 2\\ 0 & 1 & 3 & 4\\ 0 & 0 & 0 & 15\\ 0 & 0 & 0 & 0 \end{array}\right]\qquad...$ 3 pivot positions. Row 4 does not have a pivot position. In Theorem 4, $\mathrm{d}. \mathrm{B}$ has a pivot position in every row, is false. So, all statements in the theorem are false, including $\mathrm{b}$. Each $\mathrm{b}$ in $\mathbb{R}^{m}$ is a linear combination of the columns of $\mathrm{B}$. $\mathrm{c}$. The columns of $\mathrm{B}$ span $\mathbb{R}^{m}$.
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