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Linear Algebra and Its Applications 5th Edition Solutions Manual PDF

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Complete Answers Solutions Manual PDF for Linear Algebra and Its Applications 5th Edition by David C. Lay, Steven R. Lay, and Judi J. McDonald. Includes all the exercises answers for all 8 chapters.

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  • October 26, 2024
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INSTRUCTOR’S SOLUTIONS
MANUAL
JUDI J. MCDONALD
Washington State University




LINEAR ALGEBRA
AND ITS APPLICATIONS
FIFTH EDITION
GLOBAL EDITION


David C. Lay
University of Maryland


Steven R. Lay
Lee University


Judi J. McDonald
Washington State University




Boston Columbus Hoboken Indianapolis New York San Francisco
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo

,Copyright © 2016 Pearson Education Limited

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any
form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written
permission of the publisher.

Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks.
Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations
have been printed in initial caps or all caps.

ISBN-13: 978-1-292-09227-0
ISBN-10: 1-292-09227-0

, Contents



Chapter 1 1-1

Chapter 2 2-1

Chapter 3 3-1

Chapter 4 4-1

Chapter 5 5-1

Chapter 6 6-1

Chapter 7 7-1

Chapter 8 8-1




iii
Copyright © 2016 Pearson Education, Ltd.

, 1.1 SOLUTIONS

Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand
for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.

x1 + 5 x2 =
7  1 5 7
1.  −2
−2 x1 − 7 x2 =
−5  −7 −5
x1 + 5 x2 =
7 1 5 7
Replace R2 by R2 + (2)R1 and obtain: 0
3 x2 = 9  3 9 
x1 + 5 x2 =
7 1 5 7
Scale R2 by 1/3: 0
x2 = 3  1 3 
x1 = −8 1 0 −8
Replace R1 by R1 + (–5)R2: 0
x2 = 3  1 3
The solution is (x1, x2) = (–8, 3), or simply (–8, 3).

2 x1 + 4 x2 =
−4 2 4 −4 
2. 5
5 x1 + 7 x2 =
11  7 11 
x1 + 2 x2 =
−2 1 2 −2 
Scale R1 by 1/2 and obtain: 5
5 x1 + 7 x2 =
11  7 11 
x1 + 2 x2 =
−2 1 2 −2 
Replace R2 by R2 + (–5)R1: 0
−3 x2 =
21  −3 21
x1 + 2 x2 =
−2 1 2 −2 
Scale R2 by –1/3: 0
x2 = −7  1 −7 
x1 = 12 1 0 12 
Replace R1 by R1 + (–2)R2: 0
x2 = −7  1 −7 
The solution is (x1, x2) = (12, –7), or simply (12, –7).



Copyright © 2016 Pearson Education, Ltd. 1-1

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