/*! 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} Q7E In Exercises 5鈥8, use the defi... [FREE SOLUTION] | 91影视

91影视

In Exercises 5鈥8, use the definition of Ax to write the matrixequation as a vector equation, or vice versa.

7. \({x_1}\left[ {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right]\)

Short Answer

Expert verified

The vector equation \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)in a matrix equation is written as \(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

Step by step solution

01

Write the definition of \(A{\bf{x}}\)

It is known that the column of matrix \(A\) is represented as \(\left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\), and vector x is represented as \(\left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right)\).

According to the definition, the weights in a linear combination of matrix A columns are represented by the entries in vector x.

\({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\)

The left-hand side \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n}\) of the above equation is a linear combination of vectors \({x_1},{x_2},...,{x_n}\).

Thus, the vector equation in the matrix form is written as:

\(A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{ \cdot \cdot \cdot }&{{a_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\ \vdots \\{{x_n}}\end{array}} \right) = b\)

The number of columns in matrix\(A\)should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

02

Obtain the columns of the matrix

Compare the given vector equation form \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)with the general equation \({x_1}{a_1} + {x_2}{a_2} + \cdots + {x_n}{a_n} = A{\bf{x}}\) form.

So, \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), \({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\), and \(A{\bf{x}} = b = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

It shows that the equation is a linear combination of three vectors \({x_1}\), \({x_2}\), and \({x_3}\).

03

Write matrix Aand vector x

According to the definition,the number of columns in matrix\(A\)should be equal to the number of entries in vector x so that \(A{\bf{x}}\) can be defined.

Write matrix A using three columns \({{\bf{a}}_1} = \left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)\), \({{\bf{a}}_2} = \left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)\), and\({{\bf{a}}_3} = \left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)\) and vector x using three entries \({x_1}\), \({x_2}\), and \({x_3}\).

So, \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), and \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\).

04

Write the vector equation into a matrix equation

By using matrix \(A = \left( {\begin{array}{*{20}{c}}{\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right)}&{\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right)}\end{array}} \right)\), and vector \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\), the matrix equation can be written as shown below:

\(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)

Thus, the vector equation \({x_1}\left( {\begin{array}{*{20}{c}}4\\{ - 1}\\7\\{ - 4}\end{array}} \right) + {x_2}\left( {\begin{array}{*{20}{c}}{ - 5}\\3\\{ - 5}\\1\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}7\\{ - 8}\\0\\2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\)can be written as a matrix equation as \(\left( {\begin{array}{*{20}{c}}4&{ - 5}&7\\{ - 1}&3&{ - 8}\\7&{ - 5}&0\\{ - 4}&1&2\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}6\\{ - 8}\\0\\{ - 7}\end{array}} \right)\).

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

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.

Consider the problem of determining whether the following system of equations is consistent:

\(\begin{aligned}{c}{\bf{4}}{x_1} - {\bf{2}}{x_2} + {\bf{7}}{x_3} = - {\bf{5}}\\{\bf{8}}{x_1} - {\bf{3}}{x_2} + {\bf{10}}{x_3} = - {\bf{3}}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of linear combinations. Then solve that problem.
  1. Define an appropriate matrix, and restate the problem using the phrase 鈥渃olumns of A.鈥
  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

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.

In Exercises 9, write a vector equation that is equivalent to

the given system of equations.

9. \({x_2} + 5{x_3} = 0\)

\(\begin{array}{c}4{x_1} + 6{x_2} - {x_3} = 0\\ - {x_1} + 3{x_2} - 8{x_3} = 0\end{array}\)

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

18: [111315171921][-13-1]=[131921]

See all solutions

Recommended explanations on Math Textbooks

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.