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Linear Algebra and its Applications
Found in: Page 1
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

In Exercise 2, compute \(u + v\) and \(u - 2v\).

2. \(u = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\).

The vectors are \(u + v = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\) and \(u - 2v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the condition to add the two vectors

From the given vectors, it is observed that both vectors contain two entries. So, the vectors can be denoted as \({\mathbb{R}^2}\).

Add the corresponding terms of \(u\) and \(v\) to compute \(u + v\).

Step 2: Compute the sum of the vectors

Obtain vector \(u + v\) by using vectors \(u = \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right]\), and \(v = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\).

\(\begin{aligned}{c}u + v &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + \left( 2 \right)}\\{2 + \left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 + 2}\\{2 - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\end{aligned}\)

Thus, the vector is \(u + v = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\).

Step 3: Write the condition for scalar multiple of a vector by a constant

Multiply each entry of a \({\mathbb{R}^2}\) vector by a scalar number to obtain the scalar multiple of a vector by a constant.

Step 4: Compute the vector

Vector \(u - 2v\) can be written as \(u + \left( { - 2} \right)v\).

Obtain the scalar multiple of vector \(v\) by scalar \(\left( { - 2} \right)\), and then add the resultant vector with vector \(u\).

\(\begin{aligned}{c}u + \left( { - 2} \right)v &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( 2 \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}3\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 4}\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{3 - 4}\\{2 + 2}\end{array}} \right]\end{aligned}\)

Solve further to get:

\(u + \left( { - 2} \right)v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\)

Thus, \(u - 2v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\).

Therefore, \(u + v = \left[ {\begin{array}{*{20}{c}}5\\1\end{array}} \right]\), and \(u - 2v = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\).

Most popular questions for Math Textbooks

In Exercises 31, find the elementary row operation that transforms the first matrix into the second, and then find the reverse row operation that transforms the second matrix into the first.

31. \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\4&{ - 1}&3&{ - 6}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1&0\\0&5&{ - 2}&8\\0&7&{ - 1}&{ - 6}\end{array}} \right]\)

Question: There exists a 2x2 matrix such that A[12]=[34].

Suppose A is an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

Find the general solutions of the systems whose augmented matrices are given as

12. \(\left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\).

In Exercises 3 and 4, display the following vectors using arrows

on an \(xy\)-graph: u, v, \( - {\bf{v}}\), \( - 2{\bf{v}}\), u + v , u - v , and u - 2v. Notice that is the vertex of a parallelogram whose other vertices are u, 0, and \( - {\bf{v}}\).

3. u and v as in Exercise 1
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