/*! 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} Q4E In Exercise 4, find the vector x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q4E (page 191)

In Exercise 4, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

4. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{2}}\\{\bf{0}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\\{\bf{2}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{7}}}\\{\bf{3}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{8}}\\{ - {\bf{7}}}\end{array}} \right)\)

Short Answer

Expert verified

Vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\)

Step by step solution

01

Use the definition

The coordinates of x relative to basis\({\rm B} = \left\{ {{b_{\bf{1}}},{b_{\bf{2}}},...,{b_n}} \right\}\)are the weights\({c_{\bf{1}}},{c_{\bf{2}}},...,{c_n}\),such that\(x = {c_{\bf{1}}}{b_{\bf{1}}} + {c_{\bf{2}}}{b_{\bf{2}}} + ... + {c_n}{b_n}\). Then,\({\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\ \vdots \\{{c_n}}\end{array}} \right)\).

02

Find x

By the above definition, you get

\(\begin{array}{c}x = - 4\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right) + 8\left( {\begin{array}{*{20}{c}}3\\{ - 5}\\2\end{array}} \right) + \left( { - 7} \right)\left( {\begin{array}{*{20}{c}}4\\{ - 7}\\3\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}4\\{ - 8}\\0\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{24}\\{ - 40}\\{16}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - 28}\\{49}\\{ - 21}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 + 24 - 28}\\{ - 8 - 40 + 49}\\{0 + 16 - 21}\end{array}} \right)\\x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right).\end{array}\)

03

Draw a conclusion

Hence, vector \(x = \left( {\begin{array}{*{20}{c}}0\\1\\{ - 5}\end{array}} \right)\).

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

If a \({\bf{3}} \times {\bf{8}}\) matrix A has a rank 3, find dim Nul A, dim Row A, and rank \({A^T}\).

In Exercises 7-10, let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for \({\mathbb{R}^2}\). In each exercise, find the change-of-coordinates matrix from \(B\) to \(C\) and the change-of-coordinates matrix from \(C\) to \(B\).

10. \({{\mathop{\rm b}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}7\\{ - 2}\end{array}} \right),{{\mathop{\rm b}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right),{{\mathop{\rm c}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}4\\1\end{array}} \right),{{\mathop{\rm c}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right)\).

In Exercises 21 and 22, mark each statement True or False. Justify

each answer.

22. a.A linearly independent set in a subspace H is a basis for H.

b. If a finite set S of nonzero vectors spans a vector space V, then some subset of S is a basis for V.

c. A basis is a linearly independent set that is as large as possible.

d. The standard method for producing a spanning set for Nul A, described in Section 4.2, sometimes fails to produce a basis for Nul A.

e. If B is an echelon form of a matrix A, then the pivot columns of B form a basis for Col A.

Question: In Exercises 13 and 14, assume that A is row equivalent to B. Find bases for NulA and Col A.

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

A scientist solves a nonhomogeneous system of ten linear equations in twelve unknowns and finds that three of the unknowns are free variables. Can the scientist be certain that, if the right sides of the equations are changed, the new nonhomogeneous system will have a solution? Discuss.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

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.