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 2E from Chapter 1.3 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 2E
Chapter:
Problem:
In Exercises 1 and 2, compute u + v and u – 2v.
Step-by-Step Solution
Given Information
We are given with two vectors: \[{\bf{u}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right],{\bf{v}} = \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\] We have to find the entities u+v and u-2v
Step-1: u+v
\[\begin{array}{l} {\bf{u}} + {\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 + 2}\\ {2 + ( - 1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right] \end{array}\]
Step-2: u -2v
\[\begin{array}{l} {\bf{u}} - 2{\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + ( - 2)\left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {( - 2)(2)}\\ {( - 2)( - 1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 + ( - 4)}\\ {2 + 2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 4 \end{array}} \right] \end{array}\]
\[\begin{array}{l} {\bf{u}} + {\bf{v}} = \left[ {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right]\\ {\bf{u}} - 2{\bf{v}} = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 4 \end{array}} \right] \end{array}\]
We are given with two vectors: \[{\bf{u}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right],{\bf{v}} = \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\] We have to find the entities u+v and u-2v
Step-1: u+v
\[\begin{array}{l} {\bf{u}} + {\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 + 2}\\ {2 + ( - 1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right] \end{array}\]
Step-2: u -2v
\[\begin{array}{l} {\bf{u}} - 2{\bf{v}} = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + ( - 2)\left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 3\\ 2 \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {( - 2)(2)}\\ {( - 2)( - 1)} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {3 + ( - 4)}\\ {2 + 2} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 4 \end{array}} \right] \end{array}\]
\[\begin{array}{l} {\bf{u}} + {\bf{v}} = \left[ {\begin{array}{*{20}{c}} 5\\ 1 \end{array}} \right]\\ {\bf{u}} - 2{\bf{v}} = \left[ {\begin{array}{*{20}{c}} { - 1}\\ 4 \end{array}} \right] \end{array}\]