Solution Manual Linear Algebra And Its Applications
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INSTRUCTOR’S SOLUTIONS MANUAL THOMAS POLASKI Winthrop University JUDITH MCDONALD Washington State University L INEAR ALGEBRA AND I TS A PPLICATIONS F OURTH E DITION David C. Lay University of Maryland The author and publisher of this book have used their best efforts in preparing this book.
These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2012, 2006, 1997 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116.
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. Printed in the United States of America. ISBN-13: 978-0-321-38888-9 ISBN-10: 0-321-38888-7 1 2 3 4 5 6 BB 15 14 13 12 11 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 −2 x1 − 7 x2 = −5 ª 1 « −2 ¬ 5 −7 7º −5»¼ Replace R2 by R2 + (2)R1 and obtain: x1 + 5 x2 = 7 3x2 = 9 x1 + 5 x2 = 7 Scale R2 by 1/3: x2 = 3 x1 Replace R1 by R1 + (–5)R2: = −8 x2 = 3 ª1 «0 ¬ 5 ª1 «0 ¬ 5 7º 9 »¼ 3 7º 3 »¼ 1 ª1 «0 ¬ 0 ª1 «5 ¬ 2 1 −8 º 3»¼ The solution is (x1, x2) = (–8, 3), or simply (–8, 3). 3x1 + 6 x2 = −3 5 x1 + 7 x2 = 10 ª3 «5 ¬ 6 7 −3 º 10 »¼ Scale R1 by 1/3 and obtain: Replace R2 by R2 + (–5)R1: Scale R2 by –1/3: Replace R1 by R1 + (–2)R2: x1 + 2 x2 = −1 5 x1 + 7 x2 = 10 x1 + 2 x2 = −1 −3x2 = 15 x1 + 2 x2 = −1 x2 = −5 x1 = 9 x2 = −5 ª1 «0 ¬ −1º 10 »¼ 7 2 −3 ª1 «0 ¬ 2 ª1 «0 ¬ 0 1 1 −1º 15»¼ −1º −5»¼ 9º −5 »¼ The solution is (x1, x2) = (9, –5), or simply (9, –5). Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 1 1.1.
Solutions 8. The standard row operations are: ª1 «0 « «0 « ¬0 −5 4 0 1 0 0 0 3 0 1 0 2 ª1 «0 « «0 « ¬0 0º ª 1 0 »» «« 0 0» «0 » « 0¼ ¬ 0 −5 1 0 0 0 1 0 0 0 0 0 1 0º ª 1 0 »» «« 0 0» «0 » « 0¼ ¬0 −5 4 0 1 0 0 0 3 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 0º ª 1 0»» ««0 0» «0 » « 0¼ ¬0 −5 4 0 1 0 0 0 3 0 0 0 1 0º ª 1 0 »» ««0 0 » «0 » « 0¼ ¬0 −5 4 0 1 0 0 0 1 0 0 0 1 0º 0 »» 0» » 0¼ 0º 0»» 0» » 0¼ The solution set contains one solution: (0, 0, 0, 0). The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4: ª1 «0 « «0 « ¬0 −1 1 0 −2 0 0 0 0 1 0 −3 1 −5º ª 1 −7 »» «« 0 2» «0 » « 4¼ ¬0 −1 1 0 −2 0 0 0 0 1 0 0 1 −5º −7 »» 14 » » 4¼ Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2: ª1 «0 « «0 « ¬0 −1 1 0 0 0 0 0 0 1 0 0 1 −5º ª 1 21»» «« 0 14 » « 0 » « 4 ¼ ¬0 0 1 0 0 0 0 1 0 16 º 21»» 0 14 » » 1 4¼ 0 0 The solution set contains one solution: (16, 21, 14, 4).
The system has already been reduced to triangular form. Use the 1 in the fourth row to change the 3 and –2 above it to zeros. That is, replace R2 by R2 + (-3)R4 and replace R1 by R1 + (2)R4. For the final step, replace R1 by R1 + (-3)R2. ª1 «0 « «0 « ¬0 3 1 0 0 −2 3 0 0 1 0 0 1 −7 º ª 1 6 »» «« 0 2» «0 » « −2 ¼ ¬ 0 3 1 0 0 0 0 0 0 1 0 0 1 −11º ª 1 12 »» ««0 2» «0 » « −2 ¼ ¬0 0 1 0 0 0 0 0 0 1 0 0 1 −47 º 12»» 2» » −2¼ The solution set contains one solution: (–47, 12, 2, –2). First, swap R1 and R2.
Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2. ª0 «1 « ¬« 2 1 4 7 5 3 1 −4º ª 1 −2 »» «« 0 −2 ¼» ¬« 2 4 1 7 3 5 1 −2º ª 1 −4»» «« 0 −2 ¼» ¬«0 4 1 −1 3 5 −5 −2 º ª 1 −4 »» «« 0 2 ¼» ¬« 0 4 1 0 3 5 0 −2º −4»» −2¼» The system is inconsistent, because the last row would require that 0 = –2 if there were a solution. The solution set is empty.
Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 3 4 CHAPTER 1. Linear Equations in Linear Algebra 12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2.
ª 1 « 2 « «¬ −2 −5 −7 1 4 3 7 −3º ª 1 −2 »» ««0 −1»¼ «¬0 −5 3 −9 4 −5 15 −3º ª 1 4»» ««0 −7 »¼ «¬0 −5 3 0 4 −5 0 −3º 4»» 5»¼ The system is inconsistent, because the last row would require that 0 = 5 if there were a solution. The solution set is empty. «« 2 «¬ 0 ª1 «0 « «¬0 ª2 « 14.
« 0 ¬« 3 ª1 ««0 «¬0 8º ª 1 7 »» ««0 −2 »¼ «¬0 −3 9 5 0 2 1 0 1 0 −3 5 1 0 1 0 8º ª 1 −2 »» ««0 −1»¼ ¬«0 −8º ª 1 3»» ««0 −4 ¼» ¬« 3 −6 2 −2 0 1 6 −3 2 1 −4º ª 1 3»» ««0 2 »¼ «¬0 −3 15 5 0 2 1 0 1 0 0 0 1 −3 2 −2 0 1 6 0 1 0 −3 0 1 8º ª 1 −9 »» ««0 −2 »¼ «¬ 0 −3 5 15 0 1 2 8º ª 1 −2 »» ««0 −9»¼ «¬ 0 0 1 0 −3 5 5 8º −2 »» −5»¼ 0 1 0 −3 2 −5 −4º 3»» −10¼» 5º 3». The solution is (5, 3, –1).
» −1»¼ −4º ª 1 3»» ««0 −4¼» ¬«0 −4º ª 1 −1»» ««0 2 »¼ «¬0 −3 2 7 0 1 6 0 1 0 0 0 1 −4º ª 1 3»» «« 0 8¼» ¬« 0 2º −1»». The solution is (2, –1, 2). First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (– 1)R3. 0 0 5º ª 1 −6 0 0 5º ª 1 −6 0 0 5º ª 1 −6 0 0 5º ª 1 −6 « 0 » « » « » « 1 −4 1 0 » « 0 1 −4 1 0 » «0 1 −4 1 0 » « 0 1 −4 1 0 »» « « −1 6 1 5 3» « 0 0 1 5 8» «0 0 1 5 8» « 0 0 1 5 8» « » « » « » « » 5 4 0 ¼ ¬0 −1 5 4 0 ¼ ¬0 0 1 5 0¼ ¬0 0 0 0 −8¼ ¬ 0 −1 The system is inconsistent, because the last row would require that 0 = –8 if there were a solution. First replace R4 by R4 + (3/2)R1 and replace R4 by R4 + (–2/3)R2. (One could also scale R1 and R2 before adding to R4, but the arithmetic is rather easy keeping R1 and R2 unchanged.) Finally, replace R4 by R4 + (–1)R3.
ª 2 0 0 −4 −10 º ª 2 0 0 −4 −10 º ª 2 0 0 −4 −10 º ª 2 0 0 −4 −10 º « 0 3 3 0 0 »» «« 0 3 3 0 0 »» «« 0 3 3 0 0»» «« 0 3 3 0 0 »» « « 0 0 1 −1» « 0 0 1 −1» « 0 0 1 −1» « 0 0 1 −1» 4 4 4 4 « » « » « » « » −9 ¼ 1 5¼ ¬ 0 2 3 −5 −10¼ ¬ 0 0 1 −5 −10¼ ¬ 0 0 0 −9 ¬ −3 2 3 The system is now in triangular form and has a solution. In fact, using the argument from Example 2, one can see that the solution is unique. Copyright ©!2012 Pearson Education, Inc.
Publishing as Addison-Wesley. 6 CHAPTER 1. Linear Equations in Linear Algebra 7 g º ª1 −4 7 g º ª1 −4 7 g ª 1 −4 º « » « » « » h » «0 h 3 −5 h » « 0 3 −5 3 −5 25. « 0 » «¬ −2 5 −9 k »¼ «¬0 −3 5 k + 2 g »¼ «¬ 0 0 0 k + 2 g + h »¼ Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix above is 0 = b. This equation is possible if and only if b is zero.
So the original system has a solution if and only if k + 2g + h = 0. Row reduce the augmented matrix for the given system: ª2 «c ¬ 4 d f º ª1 g »¼ «¬ c 2 d f / 2 º ª1 g »¼ «¬0 2 º » g − c ( f / 2) ¼ f /2 d − 2c This shows that d – 2c must be nonzero, since f and g are arbitary. Otherwise, for some choices of f and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible since a is nonzero.
Then replace R2 by R2 + (–c)R1. ªa « ¬c b d f º ª1 g ¼» ¬« c f / a º ª1 g ¼» ¬« 0 b/a d º g − c( f / a ) ¼» b/a f /a d − c(b / a ) The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be written as ad – bc ≠ 0, or ad ≠ bc. A basic principle of this section is that row operations do not affect the solution set of a linear system.
Linear Algebra Pdf
Begin with a simple augmented matrix for which the solution is obviously (3, –2, –1), and then perform any elementary row operations to produce other augmented matrices. Here are three examples. 2006 mercury outboard service manual. The fact that they are all row equivalent proves that they all have the solution set (3, –2, – 1). ª1 «0 « ¬«0 0 1 0 0 0 1 3º ª 1 −2»» «« 2 −1¼» ¬« 0 0 1 0 0 0 1 3º ª 1 4»» «« 2 −1¼» ¬« 2 0 1 0 0 0 1 3º 4»» 5¼» 29. Swap R1 and R3; swap R1 and R3. Multiply R3 by –1/5; multiply R3 by –5. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.
Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2. The first equation was given. The others are: T2 = (T1 + 20 + 40 + T3 )/4, or 4T2 − T1 − T3 = 60 T3 = (T4 + T2 + 40 + 30)/4, or 4T3 − T4 − T2 = 70 T4 = (10 + T1 + T3 + 30)/4, or 4T4 − T1 − T3 = 40 Copyright ©!2012 Pearson Education, Inc. Publishing as Addison-Wesley.
Linear Algebra Problems
1.1. Solutions 7 Rearranging, 4T1 − T2 −T1 + 4T2 −T2 − −T1 T4 = 30 − T3 + 4T3 − T4 = 60 = 70 − + 4T4 = 40 T3 34.