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 42: 28

Answer

$\begin{bmatrix} \vec{v_1} & \vec{v_2} & \vec{v_3} & \vec{v_4} & \vec{v_5}\\ \end{bmatrix} \begin{bmatrix} c_1\\ c_2\\ c_3\\ c_4\\ c_5\\ \end{bmatrix}= \begin{bmatrix} 8\\ -1 \end{bmatrix} $ Vectors $\vec{v_1}$ to $\vec{v_5}$ and scalars $c_1$ to $c_5$ are defined below.

Work Step by Step

$\begin{bmatrix} -3 & 5 & -4 & 9 & 7\\ 5 & 8 & 1 & -2 & -4 \end{bmatrix} \begin{bmatrix} -3\\ 2\\ 4\\ -1\\ 2\\ \end{bmatrix}= \begin{bmatrix} 8\\ -1 \end{bmatrix} $ This is equivalent to $\begin{bmatrix} \vec{v_1} & \vec{v_2} & \vec{v_3} & \vec{v_4} & \vec{v_5}\\ \end{bmatrix} \begin{bmatrix} c_1\\ c_2\\ c_3\\ c_4\\ c_5\\ \end{bmatrix}= \begin{bmatrix} 8\\ -1 \end{bmatrix} $ $\vec{v_1}=\begin{bmatrix} -3\\ 5\\ \end{bmatrix}$, $\vec{v_2}=\begin{bmatrix} 5\\ 8\\ \end{bmatrix}$, $\vec{v_3}=\begin{bmatrix} -4\\ 1\\ \end{bmatrix}$, $\vec{v_4}=\begin{bmatrix} 9\\ -2\\ \end{bmatrix}$, $\vec{v_5}=\begin{bmatrix} 7\\ -4\\ \end{bmatrix}$ $c_1=-3$, $c_2=2$, $c_3=4$,$c_4=-1$, $c_5=2$
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