Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 26E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 26E
Chapter:
Problem:
Let It can be shown that 3u - 5v - w = 0. Use this fact (and no row operations) to find x1 and x2 that satisfy the equation
Step-by-Step Solution
Given Information
We are given following vectors: \[{\bf{u}} = \left[ {\begin{array}{*{20}{l}} 7\\ 2\\ 5 \end{array}} \right],{\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 3 \end{array}} \right]{\rm{ and }}{\bf{w}} = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\] Using the fact that $3u-5v-w=0$, we have to find $x_1$ and $x_2$ such that they satisfy the equation: \[\left[ {\begin{array}{*{20}{l}} 7&3\\ 2&1\\ 5&3 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\]
Step-1:
Write the given relation: \[\begin{array}{l} 3u - 5v - w = 3\left[ {\begin{array}{*{20}{l}} 7\\ 2\\ 5 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 3 \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {21}\\ 6\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}\\ 5\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {21 - 15 - 6}\\ {6 - 5 - 1}\\ {15 - 15 - 0} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\]
Step-2:
Shift the vector $w$ to the right from the equation \[\begin{array}{l} 3u - 5v - w = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ 3u - 5v = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] + w\\ \left[ {\begin{array}{*{20}{c}} u&v \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right] = w\\ \left[ {\begin{array}{*{20}{c}} u&v \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = w \end{array}\] Therefore,
\[\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right]\]
We are given following vectors: \[{\bf{u}} = \left[ {\begin{array}{*{20}{l}} 7\\ 2\\ 5 \end{array}} \right],{\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 3 \end{array}} \right]{\rm{ and }}{\bf{w}} = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\] Using the fact that $3u-5v-w=0$, we have to find $x_1$ and $x_2$ such that they satisfy the equation: \[\left[ {\begin{array}{*{20}{l}} 7&3\\ 2&1\\ 5&3 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\]
Step-1:
Write the given relation: \[\begin{array}{l} 3u - 5v - w = 3\left[ {\begin{array}{*{20}{l}} 7\\ 2\\ 5 \end{array}} \right] - 5\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 3 \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {21}\\ 6\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {15}\\ 5\\ {15} \end{array}} \right] - \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {21 - 15 - 6}\\ {6 - 5 - 1}\\ {15 - 15 - 0} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] \end{array}\]
Step-2:
Shift the vector $w$ to the right from the equation \[\begin{array}{l} 3u - 5v - w = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right]\\ 3u - 5v = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right] + w\\ \left[ {\begin{array}{*{20}{c}} u&v \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right] = w\\ \left[ {\begin{array}{*{20}{c}} u&v \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = w \end{array}\] Therefore,
\[\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 3\\ 5 \end{array}} \right]\]