• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Deutsch (DE)
Deutsch (UK)
Deutsch (US)

Americas

Europe

Answers without the blur. Sign up and see all textbooks for free! → Illustration

Q3E

Expert-verified
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

Answers without the blur.

Just sign up for free and you're in.

Register for Free I’ll do it later
Illustration

Short Answer

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

The graph of the vectors is shown below:

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write vectors u, v, \( - {\bf{v}}\), and\( - 2{\bf{v}}\)

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

Vector \( - v\) can be written as \(\left( { - 1} \right)v\).

Obtain the scalar multiple of vector \(v\) by scalar \(\left( { - 1} \right)\) to compute vector \( - v\).

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

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

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

Obtain the scalar multiple of vector \(v\) by scalar \(\left( { - 2} \right)\) to compute vector \( - 2v\).

\(\begin{aligned}{c}\left( { - 2} \right)v &= \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\end{aligned}\)

Thus, \( - 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\).

Step 2: Write vectors u + v , and u - v

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

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

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

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

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

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

Step 3: Compute vector \(u - 2v\)

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 \(u\).

\(\begin{aligned}{c}u + \left( { - 2} \right)v &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left( { - 2} \right)\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{\left( { - 2} \right)\left( { - 3} \right)}\\{\left( { - 2} \right)\left( { - 1} \right)}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 1 + 6}\\{2 + 2}\end{array}} \right]\end{aligned}\)

Solve further to get:

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

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

Step 4: Display the vectors on a graph

The graph of vectors \(u = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\), \(v = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 1}\end{array}} \right]\), \( - v = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\), \( - 2v = \left[ {\begin{array}{*{20}{c}}6\\2\end{array}} \right]\), \(u - v = \left[ {\begin{array}{*{20}{c}}2\\3\end{array}} \right]\), \(u + v = \left[ {\begin{array}{*{20}{c}}{ - 4}\\1\end{array}} \right]\), and \(u - 2v = \left[ {\begin{array}{*{20}{c}}5\\4\end{array}} \right]\) using arrows are shown below:

Thus, the graph is obtained.

Most popular questions for Math Textbooks

Determine the values(s) of \(h\) such that matrix is the augmented matrix of a consistent linear system.

18. \(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&{ - 2}\\5&h&{ - 7}\end{array}} \right]\)

A Givens rotation is a linear transformation from \({\mathbb{R}^{\bf{n}}}\) to \({\mathbb{R}^{\bf{n}}}\) used in computer programs to create a zero entry in a vector (usually a column of matrix). The standard matrix of a given rotations in \({\mathbb{R}^{\bf{2}}}\) has the form

\(\left( {\begin{aligned}{*{20}{c}}a&{ - b}\\b&a\end{aligned}} \right)\), \({a^2} + {b^2} = 1\)

Find \(a\) and \(b\) such that \(\left( {\begin{aligned}{*{20}{c}}4\\3\end{aligned}} \right)\) is rotated into \(\left( {\begin{aligned}{*{20}{c}}5\\0\end{aligned}} \right)\).

Suppose \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2}} \right\}\) is a linearly independent set in \({\mathbb{R}^n}\). Show that \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _1} + {{\mathop{\rm v}\nolimits} _2}} \right\}\) is also linearly independent.

Explain why a set \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3},{{\mathop{\rm v}\nolimits} _4}} \right\}\) in \({\mathbb{R}^5}\) must be linearly independent when \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\) is linearly independent and \({{\mathop{\rm v}\nolimits} _4}\) is not in Span \(\left\{ {{{\mathop{\rm v}\nolimits} _1},{{\mathop{\rm v}\nolimits} _2},{{\mathop{\rm v}\nolimits} _3}} \right\}\).

Suppose A is an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.
Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

Decision Maths

Read explanation

Calculus

Read explanation

Probability and Statistics

Read explanation

Statistics

Read explanation

Mechanics Maths

Read explanation

Geometry

Read explanation

Pure Maths

Read explanation

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.