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InstructorsManualLeon7edPart1

# 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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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 1 1 1 0 2 1 - 2 1 0 0 4 1 - 2 0 0 0 1 - 3 0 0 0 0 2 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 + 3 x 3 = 4 5 x 1 + 2 x 2 + 6 x 2 = - 1 (d) 4 x 1 - 3 x 2 + x 3 + 2 x 4 = 4 3 x 1 + x 2 - 5 x 3 + 6 x 4 = 5 x 1 + x 2 + 2 x 3 + 4 x 4 = 8 5 x 1 + x 2 + 3 x 3 - 2 x 4 = 7 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 first 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
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 first equation and solve for x 2 . Thus, if m 1 = m 2 , there will be a unique ordered pair ( x 1 , x 2 ) that satisfies the two equations.

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InstructorsManualLeon7edPart1 - SEVENTH EDITION LINEAR...

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