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: 25

Answer

$c_1 = -3$ $c_2 = -1$ $c_3 = 2$

Work Step by Step

Given that \begin{equation} \begin{pmatrix} 4 & -3 & 1 \\ 5 & -2 & 5 \\ -6 & 2 & -3 \end{pmatrix} \begin{pmatrix} -3 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} -7 \\ -3 \\ 10 \end{pmatrix} \end{equation} By matrix-vector multiplication, we rearrange the equation as: \begin{equation} -3*\begin{pmatrix} 4 \\ 5 \\ -6 \end{pmatrix} + -1*\begin{pmatrix} -3 \\ -2 \\ 2 \end{pmatrix} + 2*\begin{pmatrix} 1 \\ 5 \\ -3 \end{pmatrix} + = \begin{pmatrix} -7 \\ -3 \\ 10 \end{pmatrix} \end{equation} Therefore, $c_1 = -3$ $c_2 = -1$ $c_3 = 2$
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