/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q2.8-17E Determine which sets in Exercise... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q2.8-17E (page 93)

Determine which sets in Exercises 15-20 are bases for \({\mathbb{R}^{\bf{2}}}\) and \({\mathbb{R}^{\bf{3}}}\). Justify each answer.

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

Short Answer

Expert verified

The vectors are the basis for \({\mathbb{R}^3}\).

Step by step solution

01

Form a matrix using the given column vectors

The matrix formed using the column vectors is

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

02

Reduce matrix \(A\) to the upper triangular form

Interchange rows 1 and 2, i.e., \({R_1} \leftrightarrow {R_2}\).

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

At row 3, multiply row 1 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 7}&3\\0&5&6\\0&{ - 10}&{11}\end{array}} \right]\)

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \({R_3} \to {R_3} + 2{R_2}\).

\(\left[ {\begin{array}{*{20}{c}}1&{ - 7}&3\\0&5&6\\0&0&{23}\end{array}} \right]\)

03

Analyze the upper triangular matrix

The matrix has three pivots. So, A is invertible, and its column forms a basis for \({\mathbb{R}^3}\).

So, the vectors are the basis for \({\mathbb{R}^3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose P is invertible and \(A = PB{P^{ - 1}}\). Solve for Bin terms of A.

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

4. \[\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.