/*! 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} Q10E In Exercises 7鈥�10, the augment... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q10E (page 1)

In Exercises 7鈥�10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.

10. \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

The solution set of a linear system contains one solution; i.e., \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{aligned}} \right)\).

Step by step solution

01

Rewrite the given augmented matrix of a linear system

The augmented matrix of a linear system is given as

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

02

Perform elementary row operations

A basic principle states that row operations do not affect the solution set of a linear system.

To eliminate the \( - 4{x_4}\) term in the second equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Add 4 times the fourth row to the second row; i.e., \({R_2} \to {R_2} + 4{R_4}\).

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\{0 + 4\left( 0 \right)}&{1 + 4\left( 0 \right)}&{0 + 4\left( 0 \right)}&{ - 4 + 4\left( 1 \right)}&{7 + 4\left( { - 3} \right)}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

03

Eliminate the fourth element of the first row

To eliminate the \(3{x_4}\) term in the first equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&3&{ - 2}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Subtract 3 times the fourth row from the first row; i.e., \({R_1} \to {R_1} - 3{R_4}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 - 3\left( 0 \right)}&{ - 2 - 3\left( 0 \right)}&{0 - 3\left( 0 \right)}&{3 - 3\left( 1 \right)}&{ - 2 - 3\left( { - 3} \right)}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

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

04

Eliminate the second element of the first row

To eliminate the \( - 2{x_2}\) term in the first equation, perform an elementary row operationon the matrix \(\left( {\begin{aligned}{*{20}{c}}1&{ - 2}&0&0&7\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\) as shown below.

Add 2 times the second row to the first row; i.e., \({R_1} \to {R_1} + 2{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}{1 + 2\left( 0 \right)}&{ - 2 + 2\left( 1 \right)}&{0 + 2\left( 0 \right)}&{0 + 2\left( 0 \right)}&{7 + 2\left( { - 5} \right)}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

After the row operation, the matrix becomes

\(\left( {\begin{aligned}{*{20}{c}}1&0&0&0&{ - 3}\\0&1&0&0&{ - 5}\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{aligned}} \right)\)

Convert the augmented matrix into equations to get \({x_1} = - 3,\,\,{x_2} = - 5,\,\,{x_3} = 6,\,\,\) and \({x_4} = - 3\).

Hence, the required solution is \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{aligned}} \right)\).

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

In Exercises 23 and 24, key statements from this section are either quoted directly, restated slightly (but still true), or altered in some way that makes them false in some cases. Mark each statement True or False, and justify your answer. (If true, give the approximate location where a similar statement appears, or refer to a de铿乶ition or theorem. If false, give the location of a statement that has been quoted or used incorrectly, or cite an example that shows the statement is not true in all cases.) Similar true/false questions will appear in many sections of the text.

23.

a. Every elementary row operation is reversible.

b. A \(5 \times 6\)matrix has six rows.

c. The solution set of a linear system involving variables \({x_1},\,{x_2},\,{x_3},........,{x_n}\)is a list of numbers \(\left( {{s_1},\, {s_2},\,{s_3},........,{s_n}} \right)\) that makes each equation in the system a true statement when the values \ ({s_1},\, {s_2},\, {s_3},........,{s_n}\) are substituted for \({x_1},\,{x_2},\,{x_3},........,{x_n}\), respectively.

d. Two fundamental questions about a linear system involve existence and uniqueness.

Suppose Ais 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.

Use the accompanying figure to write each vector listed in Exercises 7 and 8 as a linear combination of u and v. Is every vector in \({\mathbb{R}^2}\) a linear combination of u and v?

7.Vectors a, b, c, and d

Question: Determine whether the statements that follow are true or false, and justify your answer.

16: There exists a 2x2 matrix such that

A[11]=[12]andA[22]=[21].

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.

30.\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&{ - 2}&6\\0&{ - 5}&9\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}1&3&{ - 4}\\0&1&{ - 3}\\0&{ - 5}&9\end{array}} \right]\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

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.