InstructorsManualLeon7edPart1 - SEVENTH EDITION LINEAR...

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SEVENTH EDITION LINEAR ALGEBRA WITH APPLICATIONS Instructor’s Solutions Manual Steven J. Leon
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PREFACE This solutions manual is designed to accompany the seventh edition of Linear Algebra with Applications by Steven J. Leon. The answers in this manual supple- ment those given in the answer key of the textbook. In addition this manual contains the complete solutions to all of the nonroutine exercises in the book. At the end of each chapter of the textbook there are two chapter tests (A and B) and a section of computer exercises to be solved using MATLAB. The questions in each Chapter Test A are to be answered as either true or false . Although the true- false answers are given in the Answer Section of the textbook, students are required to explain or prove their answers. This manual includes explanations, proofs, and counterexamples for all Chapter Test A questions. The chapter tests labelled B contain workout problems. The answers to these problems are not given in the Answers to Selected Exercises Section of the textbook, however, they are provided in this manual. Complete solutions are given for all of the nonroutine Chapter Test B exercises. In the MATLAB exercises most of the computations are straightforward. Con- sequently they have not been included in this solutions manual. On the other hand, the text also includes questions related to the computations. The purpose of the questions is to emphasize the signiFcance of the computations. The solutions man- ual does provide the answers to most of these questions. There are some questions for which it is not possible to provide a single answer. ±or example, aome exercises involve randomly generated matrices. In these cases the answers may depend on the particular random matrices that were generated. Steven J. Leon sleon@umassd.edu
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TABLE OF CONTENTS Chapter 1 1 Chapter 2 26 Chapter 3 37 Chapter 4 63 Chapter 5 75 Chapter 6 110 Chapter 7 150
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CHAPTER 1 SECTION 1 2. (d) 1 1 111 021 - 21 004 1 - 2 000 1 - 3 0 0 002 5. (a) 3 x 1 +2 x 2 =8 x 1 +5 x 2 =7 (b) 5 x 1 - 2 x 2 + x 3 =3 2 x 1 +3 x 2 - 4 x 3 =0 (c) 2 x 1 + x 2 +4 x 3 = - 1 4 x 1 - 2 x 2 x 3 =4 5 x 1 x 2 +6 x 2 = - 1 (d) 4 x 1 - 3 x 2 + x 3 x 4 3 x 1 + x 2 - 5 x 3 x 4 =5 x 1 + x 2 x 3 x 4 5 x 1 + x 2 x 3 - 2 x 4 9. Given the system - m 1 x 1 + x 2 = b 1 - m 2 x 1 + x 2 = b 2 one can eliminate the variable x 2 by subtracting the frst row From the second. One then obtains the equivalent system - m 1 x 1 + x 2 = b 1 ( m 1 - m 2 ) x 1 = b 2 - b 1 1
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2 CHAPTER 1 (a) If m 1 ± = m 2 , then one can solve the second equation for x 1 x 1 = b 2 - b 1 m 1 - m 2 One can then plug this value of x 1 into the Frst equation and solve for x 2 . Thus, if m 1 ± = m 2 , there will be a unique ordered pair ( x 1 ,x 2 ) that satisFes the two equations.
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InstructorsManualLeon7edPart1 - SEVENTH EDITION LINEAR...

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