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
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
