(Solution Manual) Instructor Solutions Manual for Linear Algebra and Its Applications

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INSTRUCTOR ’S SOLUTIONS MANUAL THOMAS POLASKI Winthrop University JUDITH MCDONALD Washington State University LINEAR ALGEBRA AND ITS APPLICATIONS FOURTH EDITION David C. Lay University of Maryland The author and publisher of this book have use d 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 regar d 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 pu blication 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 Am erica. 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 iii _____________________________________________________ Contents CHAPTER 1 Linear Equations in Linear Algebra 1 CHAPTER 2 Matrix Algebra 87 CHAPTER 3 Determinants 167 CHAPTER 4 Vector Spaces 197 CHAPTER 5 Eigenvalues and Eigenvectors 273 CHAPTER 6 Orthogonality and Least Squares 357 CHAPTER 7 Symmetric Matrices and Quadratic Forms 405 CHAPTER 8 The Geometry of Vector Spaces 453 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 1 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. 1. 12 1257 27 5xx xx+= −− =− 157 275ª º « »−−−¬ ¼ Replace R2 by R2 + (2)R1 and obtain: 12 257 39xx x+= = 157 039ª º « »¬ ¼ Scale R2 by 1/3: 12 257 3xx x+= = 157 013ª º « »¬ ¼ Replace R1 by R1 + (–5)R2: 1 28 3x x=− = 10 8 01 3−ª º « »¬ ¼ T h e s o l u t i o n i s ( x1, x2) = (–8, 3), or simply (–8, 3). 2. 12 1236 3 57 1 0xx xx+= − += 36 3 571 0−ª º « »¬ ¼ Scale R1 by 1/3 and obtain: 12 1221 57 1 0xx xx+= − += 12 1 571 0−ª º « »¬ ¼ Replace R2 by R2 + (–5)R1: 12 221 31 5xx x+= − − = 121 03 1 5− ª º « »−¬ ¼ Scale R2 by –1/3: 12 221 5xx x+= − =− 12 1 01 5− ª º « »−¬ ¼ Replace R1 by R1 + (–2)R2: 1 29 5x x= =− 10 9 01 5ª º « »−¬ ¼ T h e s o l u t i o n i s ( x1, x2) = (9, –5), or simply (9, –5). 2 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 3. The point of intersection satisfies the system of two linear equations: 12 1224 1xx xx+= − = 12 4 11 1ªº «»−¬¼ Replace R2 by R2 + (–1)R1 and obtain: 12 224 33xx x+= − =− 124 033ª º « »−−¬ ¼ Scale R2 by –1/3: 12 224 1xx x+= = 124 011ª º « »¬ ¼ Replace R1 by R1 + (–2)R2: 1 22 1x x= = 102 011ª º « »¬ ¼ The point of intersection is ( x1, x2) = (2, 1). 4. The point of intersection satisfies the system of two linear equations: 12 1221 3 32 1xx xx+= − − = 121 3 32 1− ªº «»−¬¼ Replace R2 by R2 + (–3)R1 and obtain: 12 221 3 84 0xx x+= − − = 121 3 084 0− ª º « »−¬ ¼ Scale R2 by –1/8: 12 221 3 5xx x+= − =− 12 1 3 01 5−ª º « »− ¬ ¼ Replace R1 by R1 + (–2)R2: 1 23 5x x=− =− 10 3 01 5−ª º « »−¬ ¼ The point of intersection is ( x1, x2) = (–3, –5). 5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with –4 times R3, and then replace R1 by its sum with 3 times R3. 6. One more step will put the system in triangular form. Replace R4 by its sum with –4 times R3, which produces 16401 02704 00123 0007 1 4−−ªº «»−«» «» − «»− ¬¼. After that, the next step is to scale the fourth row by –1/7. 7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0 x2 + 0 x 3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty. 1.1 • Solutions 3 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 8. The standard row operations are: 15 4 0 0 15 4 0 0 15 4 0 0 15 4 0 0 01 0 1 0 01 0 1 0 01 0 0 0 01 0 0 0~~~00 3 0 0 00 3 0 0 00 3 0 0 00 1 0 0 00 0 2 0 00 0 1 0 00 0 1 0 00 0 1 0 15 0 0 0 1 0 0 0 0 01 0 0 0 0 1 0 0 0~~00 1 0 0 0 0 1 0 0 00 0 1 0 0 0 0 1 0−−−−ªº ªº ªº ªº «» «» «» «» «» «» «» «» «» «» «» «» «» «» «» «» ¬¼ ¬¼ ¬¼ ¬¼ −ªº ªº «» «» «» «» «» « «» « ¬¼ ¬¼» » The solution set contains one solution: (0, 0, 0, 0). 9. The system has already been reduced to triangular form. Begin by replacing R3 by R3 + (3)R4: 11005 110 05 01207 012 07~00 132 00 1 0 1 4 00014 000 14−− − −ªº ª º «» « »−− − −«» « » «» « » − «» « » ¬¼ ¬ ¼ Next, replace R2 by R2 + (2)R3. Finally, replace R1 by R1 + R2: 11 0 05 1 0 0 0 1 6 01 0 0 2 1 0 1 0 0 2 1~~00 1 0 1 4 0 0 1 0 1 4 00 0 14 0 0 0 14−−ªº ª º «» « » «» « » «» « » «» « » ¬¼ ¬ ¼ The solution set contains one solution: (16, 21, 14, 4). 10. 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. 130 2 7 1300 1 1 1000 4 7 01 0 3 6 01 00 1 2 01 00 1 2~~001 0 2 0010 2 0010 2 000 1 2 0001 2 0001 2−− − − ªº ªº ªº «» «» «» «» «» «» «» «» «» «» «» «»−−− ¬¼ ¬¼ ¬¼ The solution set contains one solution: (–47, 12, 2, –2). 11. First, swap R1 and R2. Then replace R3 by R3 + (–2)R1. Finally, replace R3 by R3 + (1)R2. 01 5 4 1 43 2 1 4 3 2 1 43 2 143 2 ~ 0 15 4 ~ 0 1 5 4 ~ 0 15 4 271 2 271 2 0 1 5 2 000 2−− −− ªº ªº ª º ªº «» «» « » «»−− −−«» «» « » «» «» «» « » «» −− − −− ¬¼ ¬¼ ¬ ¼ ¬¼ The system is inconsistent, because the last row would require that 0 = –2 if there were a solution. The solution set is empty. 4 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 12. Replace R2 by R2 + (–2)R1 and replace R3 by R3 + (2)R1. Finally, replace R3 by R3 + (3)R2. 15 43 1543 1543 27 32 ~ 0354 ~ 0354 21 7109 1 57 0005−− − − − −ªº ªº ªº «» «» «»−− − −«» «» «» «» «» «» −− − −¬¼ ¬¼ ¬¼ The system is inconsistent, because the last row would require that 0 = 5 if there were a solution. The solution set is empty. 13. 10 3 8 10 3 8 10 3 8 10 3 8 22 9 7 ~ 021 5 9 ~ 01 5 2 ~ 01 5 2 01 5 2 01 5 2 021 5 9 00 5 5−−−−ªº ªº ªº ªº «» «» «» «»−−−«» «» «» «» «» «» «» «» −−−− ¬¼ ¬¼ ¬¼ ¬¼ 10 3 8 100 5 ~0 1 5 2~0 1 0 3 00 1 1 001 1−ªº ª º «» « »−«» « » «» « » −− ¬¼ ¬ ¼. The solution is (5, 3, –1). 14. 20 6 8 10 3 4 10 3 4 10 3 4 01 2 3 ~ 01 2 3 ~ 01 2 3 ~ 01 2 3 36 2 4 36 2 4 06 7 8 00 5 1 0−− −− −− − −ªº ªº ªº ªº «» «» «» «» «» «» «» «» «» «» «» «» −− −− −− ¬¼ ¬¼ ¬¼ ¬¼ 10 3 4 10 3 4 100 2 ~0 1 2 3~0 1 0 1~0 1 0 1 . 00 1 2 00 1 2 001 2−− −−ªº ªº ªº «» «» «»−−«» «» «» «» «» «»¬¼ ¬¼ ¬¼ The solution is (2, –1, 2). 15. First, replace R3 by R3 + (1)R1, then replace R4 by R4 + (1)R2, and finally replace R4 by R4 + (– 1)R3. 160 0 5 160 0 5 160 0 5 160 05 01 4 1 0 01 4 1 001 4 1 0 01 4 1 0~~~16 1 5 3 00 1 5 8 00 1 5 8 00 1 58 01 5 4 0 01 5 4 0001 5 0 000 08−−−−ªº ªº ªº ªº «» «» «» «»−−−−«» «» «» «» «» «» «» «» − «» «» «» «»−− − ¬¼ ¬¼ ¬¼ ¬¼ The system is inconsistent, because the last row would require that 0 = –8 if there were a solution. 16. 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. 200 4 1 0 200 4 1 0 200 4 1 0 200 4 1 0 033 0 0 033 0 0 033 0 0 033 0 0~~~001 4 1 001 4 1 001 4 1 001 4 1 323 1 5 023 5 1 0 001 5 1 0 000 9 9−− −− −− −− ªº ªº ªº ªº «» «» «» «» «» «» «» «» «» «» «» «» −−−− «» «» «» «»−− − − − − − ¬¼ ¬¼ ¬¼ ¬¼ 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. 1.1 • Solutions 5 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 17. Row reduce the augmented matrix corresponding to the given system of three equations: 231231231 650 ~ 043 ~ 043257 088 002−−− ªº ªº ªº «» «» «»−−«» «» «» «» «» «»−−¬¼ ¬¼ ¬¼ The third equation, 0 = 2, shows that the system is inconsistent, so the three lines have no point in common. 18. Row reduce the augmented matrix corresponding to the given system of three equations: 24 4 4 2 4 4 4 24 4 4 01 2 2 ~ 0 1 2 2 ~ 01 2 223 0 0 0 1 4 4 00 6 6ªº ª º ªº «» « » «»−− −− −−«» « » «» «» « » «» −−− −− ¬¼ ¬ ¼ ¬¼ The system is consistent, and using the argument from Example 2, there is only one solution. So the three planes have only one point in common. 19. 14 1 4~368 06 3 4hh hªº ª º «» « »−− ¬¼ ¬ ¼ Write c for 6 – 3 h. If c = 0, that is, if h = 2, then the system has no solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution. 20. 15 1 5~286 08 21 6hh h−− ªº ª º «» « »−− − ¬¼ ¬ ¼ Write c for 82 . h −− If c = 0, that is, if h = –4, then the system has no solution, because 0 cannot equal 16. Otherwise, when h ≠ –4, the system has a solution. 21. 14 2 1 4 2~36 0 1 2 0hh−−ªº ª º«» « » −−¬¼ ¬ ¼ Write c for 12h−. Then the second equation cx 2 = 0 has a solution for every value of c. So the system is consistent for all h. 22. 41 241 2~263 00 32hh h−ªº−ªº«»«»«» −− −+ ¬¼¬¼ The system is consistent if and only if 32h−+ = 0, that is, if and only if h = 6. 23. a. True. See the remarks following the box titled Elementary Row Operations. b. False. A 5 × 6 matrix has five rows. c. False. The description applies to a single solution. The solution set consists of all possible solutions. Only in special cases does the solution set consist of exactly one solution. Mark a statement True only if the statement is always true. d. True. See the box before Example 2. 24. a . False. The definition of row equivalent requires that there exist a sequence of row operations that transforms one matrix into the other. b. True. See the box preceding the subsection titled Existence and Uniqueness Questions. c. False. The definition of equivalent systems is in the second paragraph after equation (2). d. True. By definition, a consistent system has at least one solution. 6 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 25. 147 147 147 035 ~ 035 ~ 035 259 035 2 000 2gg g hh h kk gk gh−− −ªº ª º ª º «» « » « »−− −«» « » « » «» « » « » −− − ++ + ¬¼ ¬ ¼ ¬ ¼ L e t 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. 26. Row reduce the augmented matrix for the given system: 24 12 / 2 1 2 / 2~~02 ( / 2 )ff f cd g cd g d cg c fªº ª º ª º «» « » « »−− ¬¼ ¬ ¼ ¬ ¼ This shows that d – 2 c 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. Thus d 2c. 27. 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. 1/ / 1 / /~~0( / )( / )abf b af a b a f a cd g c d g d c b a g c f aªº ª º ª º «» « » « »−− ¬¼ ¬ ¼ ¬ ¼ T h e q u a n t i t y 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. 28. A basic principle of this section is that row operations do not affect the solution set of a linear system. 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. The fact that they are all row equivalent proves that they all have the solution set (3, –2, – 1). 100 3 100 3 1003 01 0 2 ~ 21 0 4 ~ 21 04 001 1 001 1 2015ªº ªº ª º «» «» « »−«» «» « » «» «» « » −− ¬¼ ¬¼ ¬ ¼ 29. Swap R1 and R3; swap R1 and R3. 30. Multiply R3 by –1/5; multiply R3 by –5. 31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1. 32. Replace R3 by R3 + (–4)R2; replace R3 by R3 + (4)R2. 33. The first equation was given. The others are: 21 3 213 (2 0 4 0) / 4 ,o r 4 6 0 TT T T T T=+ + + −− = 34 2 342 (4 0 3 0 ) / 4 , o r 4 70 TT T TTT =+ + + −− = 41 3 413 (10 30) / 4, or 4 40 TT T T T T=+ + + −− = 1.1 • Solutions 7 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. R e a r r a n g i n g , 12 4 123 234 13 443 0 46 0 47 0 44 0TT T TTT TTT TT T−− = − + − = − + − = −− += 34. Begin by interchanging R1 and R4, then create zeros in the first column: 41 01 3 0 1014 4 0 1 014 4 0 14 10 6 0 1410 6 0 04 0 42 0~~01 41 7 0 0141 7 0 0141 7 0 10 14 4 0 4101 3 0 0 14 1 5 1 9 0−− −− −−ªº ªº ªº «» «» «»−− −− −«» «» «» «» «» «» −− −− − − «» «» «»−− −− − −¬¼ ¬¼ ¬¼ Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3: 10 14 4 0 1 0 14 4 0 1 0 14 4 0 0101 5 0 101 5 0 1 01 5~~ ~01417 0 0 0427 5 0 0 427 5 014 1 5 1 9 0 0 04 1 4 1 9 5 0 0 0 1 2 2 7 0−− −− −− ªº ªº ªº «» «» «»−− −«» «» «» «» «» «» −− − − «» «» «»−− − ¬¼ ¬¼ ¬¼ Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 10 1 4 4 0 10 10 5 0 10 10 5 0 01 0 1 5 01 002 7 . 5 01 002 7 . 5~~ ~004 2 7 5 0040 1 2 0 0010 3 0 000 12 2 . 5 00012 2 . 5 00012 2 . 5−− ªº ªº ªº «» «» «»−«» «» «» «» «» «» − «» «» «» ¬¼ ¬¼ ¬¼ The last step is to replace R1 by R1 + (–1)R3: 10002 0 . 0 01 002 7 . 5~.00103 0 . 0 00012 2 . 5ªº «» «» «»«» ¬¼ The solution is (20, 27.5, 30, 22.5). Notes : The Study Guide includes a “Mathematical Note” about statements, “If … , then … .” This early in the course, students typically use single row operations to reduce a matrix. As a result, even the small grid for Exercise 34 leads to about 80 multiplications or additions (not counting operations with zero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization. For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/84+/89 calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section 1.1 describes how to access the data that is available for all numerical exercises in the text. This feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale. These commands are included in the text data sets, available from the text web site, www.pearsonhighered.com/lay. 8 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 1.2 SOLUTIONS Notes : The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises 7–14, in preparation for Section 1.5.) 1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c. 2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c. 3. 124 8 12 4 8 12 4 8 246 8 ~ 00 2 8 ~ 00 1 4 3691 2 00 3 1 2 00 3 1 2ªº ª º ª º «» « » « »−−«» « » « » «» « » « » −− −− ¬¼ ¬ ¼ ¬ ¼ 1248 120 8 ~0 0 1 4~0 0 1 4 0000 000 0− ªº ª º «» « »«» « » «» « »¬¼ ¬ ¼. Pivot cols 1 and 3. 124 8 246 8 3691 2ª º « » « » « »¬ ¼ 4. 1245 1 2 4 5 1 2 4 5 1 2 4 5 2454 ~ 0 0 3 6 ~ 0 3 1 2 1 8 ~ 0 1 4 6 4542 0 3 1 2 1 8 0 0 3 6 0 0 3 6ªº ª º ª º ª º «» « » « » « »−− −− −«» « » « » « » «» « » « » « » −− − − − −− ¬¼ ¬ ¼ ¬ ¼ ¬ ¼ 1245 124 5 100 1 ~0 1 4 6~0 1 0 2~0 1 0 2 0012 001 2 001 2ªº ª º ª º «» « » « »−−«» « » « » «» « » « »¬¼ ¬ ¼ ¬ ¼. Pivot cols 1, 2, and 3 1245 2454 4542ªº «»«» «»¬¼ 5. ** 0,,00 0 0 0ªº ªº ªº «» «» «»¬¼ ¬¼ ¬¼„„ „ „ 6.** 0 0, 0 0 , 0 0 00 0000ªºª ºª º «»« »« » «»« »« » «»« »« »¬¼¬ ¼¬ ¼„„ „ „ 7. 1347 13 4 7 1347 130 5~~ ~3976 00 5 1 5 00 13 00 1 3− ªº ªº ªº ªº «» «» «» «»−− ¬¼ ¬¼ ¬¼ ¬¼ Corresponding system of equations: 12 335 3xx x+= − = The basic variables (corresponding to the pivot positions) are x1 and x 3. The remaining variable x2 is free. Solve for the basic variables in terms of the free variable. The general solution is 12 2 353 is free 3xx xx=−−­ °® °=¯ Note : Exercise 7 is paired with Exercise 10. 1.2 • Solutions 9 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 8. 13 0 51 3 0 51 3 0 51 0 0 4~~ ~37 0 9 0 2 0 6 0 1 0 3 0 1 0 3−− − − − −ªº ªº ªº ªº «» «» «» «»−− − ¬¼ ¬¼ ¬¼ ¬¼ Corresponding system of equations: 1 24 3x x= = The basic variables (corresponding to the pivot positions) are x1 and x 2. The remaining variable x3 is free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic variables do not depend on the value of the free variable. General solution: 1 2 34 3 is freex xx=­ °=® °¯ Note : A c o m m o n e r r o r i n E x e r c i s e 8 i s t o a s s u m e t h a t x3 i s z e r o . T o a v o i d t h i s , i d e n t i f y t h e b a s i c variables first. Any remaining variables are free . (This type of computation will arise in Chapter 5.) 9. 0123 1346 1 02 3~~1346 0 123 0 12 3−− − − ªº ªº ªº «» «» «»−− − −¬¼ ¬¼ ¬¼ Corresponding system: 13 2323 23xx xx− = − = B a s i c v a r i a b l e s : x1, x2; free variable: x3. General solution: 13 23 332 32 is freexx xxx=+­ °=+® °¯ 10. 12 1 4 12 1 4 12 02~~245 6 007 1 4 00 12−− −− −ªº ªº ªº «» «» «»−− − − ¬¼ ¬¼ ¬¼ Corresponding system: 12 322 2xx x− = =− B a s i c v a r i a b l e s : x1, x3; free variable: x2. General solution: 12 2 322 is free 2xx xx=+­ °® °=−¯ 11. 324 0 3240 12 3 4 3 0 96 1 2 0 ~ 0000 ~ 0 0 0 0 64 8 0 0000 0 0 0 0−− −ªº ª º ª º «» « » « »−«» « » « » «» « » « » −¬¼ ¬ ¼ ¬ ¼ Corresponding system: 12324033 00 00xx x− += == 10 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. B a s i c v a r i a b l e : x1; free variables x2, x3. General solution: 123 2 324 33 is free is freexx x x x­= −° ° ® ° ° ¯ 12. Since the bottom row of the matrix is equivalent to the equation 0 = 1, the system has no solutions. 13. 130102 13 0 09 2 1 0 0 03 5 01 004101 0 04 10 1 0 04 1~~000194 00 0 19 4 0 0 0 19 4 000000 00 0 00 0 0 0 0 00 0−−− − − ªº ªº ªº «» «» «»−− −«» «» «» «» «» «» «» «» «» ¬¼ ¬¼ ¬¼ Corresponding system: 15 25 4535 41 94 00xx xx xx− = − = += = B a s i c v a r i a b l e s : x1, x2, x4; free variables: x3, x5. General solution: 15 25 3 45 553 14 is free 49 is freexx xxxxxx=+­ °=+°°® °=−°°¯ Note : The Study Guide discusses the common mistake x 3 = 0. 14. 10 5 0 8 3 10 5 00 3 014 1 0 6 014 1 0 6~00 0 0 1 0 00 0 01 0 00 0 0 0 0 00 0 00 0−− − ªº ªº «» «»−−«» «» «» «» «» «» ¬¼ ¬¼ Corresponding system: 13 23 4 553 46 0 00xx xx x x− = + − = == B a s i c v a r i a b l e s : x1, x2, x5; free variables: x3, x4. General solution: 13 23 4 3 4 535 64 is free is free 0xxxx xx x x=+­ °=− +°°® °° =°¯ 15. a. The system is consistent. There are many solutions because x3 is a free variable. b. The system is consistent. There are many solutions because x1 is a free variable. 16. a. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column). 1.2 • Solutions 11 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. b. The system is consistent. There are many solutions because x2 is a free variable. 17. 11 4 11 4~23 0 1 8 hh−−ªº ª º «» « »− + ¬¼ ¬ ¼ The system has a solution for all values of h since the augmented column cannot be a pivot column. 18. 131 1 3 1~62 0 3 6 2 hh h−−ªº ª º«» « » − + −− ¬¼ ¬ ¼ If 3h + 6 is zero, that is, if h = –2, then the system has a solution, because 0 equals 0. When 2, h≠− the system has a solution since the augmented column cannot be a pivot column. Thus the system has a solution for all values of h. 19. 12 1 2~48 08 4 8hh kh kªº ª º «» « »−− ¬¼ ¬ ¼ a. When h = 2 and 8,k≠ the augmented column is a pivot column, and the system is inconsistent. b. When 2,h≠ the system is consistent and has a unique solution. There are no free variables. c. When h = 2 and k = 8, the system is consistent and has many solutions. 20. 13 1 1 3 1~20 6 2 hk h k−−ªº ª º «» « »+ − ¬¼ ¬ ¼ a. When h = –6 and 2, k≠ the system is inconsistent, because the augmented column is a pivot column. b. When 6, h≠− the system is consistent and has a unique solution. There are no free variables. c. When h = –6 and k = 2, the system is consistent and has many solutions. 21. a. False. See Theorem 1. b. False. See the second paragraph of the section. c. True. Basic variables are defined after equation (4). d. True. This statement is at the beginning of Parametric Descriptions of Solution Sets . e. False. The row shown corresponds to the equation 5 x4 = 0, which does not by itself lead to a contradiction. So the system might be consistent or it might be inconsistent. 22. a. True. See Theorem 1. b. False. See Theorem 2. c. False. See the beginning of the subsection Pivot Positions . The pivot positions in a matrix are determined completely by the positions of the leading entries in the nonzero rows of any echelon form obtained from the matrix. d. True. See the paragraph just before Example 4. e. False. The existence of at least one solution is not related to the presence or absence of free variables. If the system is inconsistent, the solution set is empty. See the solution of Practice Problem 2. 12 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 23. Since there are four pivots (one in each row), the augmented matrix must reduce to the form 1 2 3 41000 0100 and so 0010 0001xa a xb b xc c xd d= ªº «»=«» «» = «»= ¬¼ No matter what the values of a, b, c, and d, the solution exists and is unique. 24. The system is consistent because there is not a pivot in column 5, which means that there is not a row of the form [0 0 0 0 1]. Since the matrix is the augmented matrix for a system, Theorem 2 shows that the system has a solution. 25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom row, and there is no room for a pivot in the augmented column. So, the system is consistent, by Theorem 2. 26. Since the coefficient matrix has three pivot columns, there is a pivot in each row of the coefficient matrix. Thus the augmented matrix will not have a row of the form [0 0 0 0 0 1], and the system is consistent. 27. “If a linear system is consistent, then the solution is unique if and only if every column in the coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ” This statement is true because the free variables correspond to nonpivot columns of the coefficient matrix. The columns are all pivot columns if and only if there are no free variables. And there are no free variables if and only if the solution is unique, by Theorem 2. 28. Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column. 29. An underdetermined system always has more variables than equations. There cannot be more basic variables than there are equations, so there must be at least one free variable. Such a variable may be assigned infinitely many different values. If the system is consistent, each different value of a free variable will produce a different solution, and the system will not have a unique solution. If the system is inconsistent, it will not have any solution. 30. Example: 123 1234 22 25xxx xxx++=++= 31. Yes, a system of linear equations with more equations than unknowns can be consistent. Example (in which x1 = x 2 = 1): 12 12 122 0 32 5xxxx xx+= − = += 32. According to the numerical note in Section 1.2, when n = 20 the reduction to echelon form takes about 2(20)3/3 5,333 flops, while further reduction to reduced echelon form needs at most (20)2 = 400 flops. Of the total flops, the “backward phase” is about 400/5733 = .07 or about 7%. When n = 200, the estimates are 2(200)3/3 5,333,333 flops for the reduction to echelon form and (200)2 = 40,000 flops for the backward phase. The fraction associated with the backward phase is about (4×104) /(5.3 ×106) = .007, or about .7%. 1.2 • Solutions 13 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 33. For a quadratic polynomial p(t) = a 0 + a 1t + a 2t2 to exactly fit the data (1, 6), (2, 15), and (3, 28), the coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the augmented matrix: 111 6 111 6 1116 1116 1241 5 ~ 0 13 9 ~ 0 139 ~ 0 1391392 8 0282 2 0024 0012ªº ªº ª º ª º «» «» « » « »«» «» « » « » «» «» « » « »¬¼ ¬¼ ¬ ¼ ¬ ¼ 1104 1001 ~0 1 0 3~0 1 0 3 0012 0012ªº ªº «» «»«» «» «» «»¬¼ ¬¼ T h e p o l y n o m i a l i s p(t) = 1 + 3t + 2t2. 34. [M] The system of equations to be solved is: 2345 01 2 3 4 5 2345 01 2 3 4 5 2345 01 2 3 4 5 2345 01 2 3 4 5 2345 01 2 3 4 5 23 01 2 300 0 0 0 0 22 2 2 2 2 . 9 0 4 4 4 4 4 14.8 6 6 6 6 6 39.6 8 8 8 8 8 74.3 10 10 10aa a a a a aa a a a a aa a a a a aa a a a aaa a a a aaa a a+ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = + ⋅ + ⋅ + ⋅ + ⋅ + ⋅ = + ⋅ + ⋅ + ⋅ + 45 4510 10 119aa⋅ + ⋅ = The unknowns are a0, a1, …, a 5. Use technology to compute the reduced echelon of the augmented matrix: 23 4 510 0 0 0 0 0 10 0 0 0 0 0 12 4 8 1 6 3 2 2 . 9 02 4 8 1 6 3 2 2 . 9 141 6 6 4 2 5 6 1 0 2 4 1 4 . 8 00 8 4 8 2 2 4 9 6 0 9~163 6 2 1 6 1 2 9 6 7 7 7 6 3 9 . 6 002 41 9 21 2 4 8 7 6 8 0 3 0 . 9 186 4 5 1 2 4 0 9 6 3 2 7 6 8 7 4 . 3 004 84 8 04 0 3 23 2 6 4 0 6 2 . 7 008 09 6 09 9 2 09 9 8 4 01 0 11 01 0 1 0 1 0 1 0 1 1 9ªº «» «»«»«»«»«» «» «»¬¼4.5ª º « » « » « » « » « » « » « »« » ¬ ¼ 100 0 0 0 0 100 0 0 0 0 024 8 1 6 3 2 2 . 9 024 8 1 6 3 2 2 . 9 008 4 8 2 2 4 9 6 0 9 0084 8 2 2 4 9 6 0 9~~000 4 8 5 7 6 4 8 0 0 3 . 9 0004 8 5 7 6 4 8 0 0 3 . 90001 9 22 6 8 82 6 8 8 0 8 . 7 000 0 3 8 4 7 6 8 0 6 . 90004 8 07 6 8 09 0 2 4 01 4 . 5 000 01 9 2 04 2 2 4 0 2 4 . 5ªº ªº «» «» «» «» «» «» «» «» «» «» «» «» −«» «− «» « ¬¼ ¬¼»» 14 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 100 0 0 0 0 100 0 0 0 0 024 8 1 6 3 2 2 . 9 024 8 1 6 3 2 2 . 9 0084 82 2 4 9 6 0 9 0084 82 2 4 9 6 0 9~~0004 85 7 64800 3.9 0004 85 7 64 8 0 0 3 . 9 000 03 8 47 6 8 0 6 . 9 000 03 8 47 6 8 0 6 . 9 000 0 03 8 4 0 1 0 000 0 0 1 . 0 0 2 6ªº ª º «» « » «» « » «» « » «» « » «» « » «» « » −−«» « »«» « » ¬¼ ¬ ¼ 100 0 00 0 100000 0 024 8 1 60 2 . 8 1 6 7 010000 1 . 7 1 2 5 0084 82 2 40 6 . 5 0 0 0 001000 1 . 1 9 4 8~~ ~0004 85 7 60 8 . 6000 0 0 0 1 0 0 .6615 000 03 8 40 2 6 . 9 0 0 000010 . 0 7 0 1000 0 01. 0 0 2 6 0 4 000001 . 0 0 2 6ªº ª º «» « » «» « » «» « » − «» « »−«» « » «» « » −−«» « » «» « » ¬¼ ¬ ¼" T h u s p(t) = 1.7125 t – 1.1948t2 + .6615 t3 – .0701 t4 + .0026 t5, and p(7.5) = 64.6 hundred lb. Notes : In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) = 64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists. Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2. The Study Guide includes a “Mathematical Note” about the phrase, “If and only if,” used in Theorem 2. 1.3 SOLUTIONS Notes : The key exercises are 11–16, 19–22, 25, and 26. A discussion of Exercise 25 will help students understand the notation [ a1 a2 a 3], {a1, a2, a3}, and Span{ a1, a2, a3}. 1. 131 ( 3 ) 4 212 ( 1 )1−−− +−− ªºªºª ºªº+= + = =«»«»« »«»− +− ¬¼¬¼¬ ¼¬¼uv . Using the definitions carefully, 13 1 ( 2 ) ( 3 ) 1 6 52( 2 )21 2 ( 2 ) ( 1 ) 2 2 4−− − − − − + ªº ªºªºª ºª ºª º− =+ − =+ = = «» «»«»« »« »« »−− − + ¬¼ ¬¼¬¼¬ ¼¬ ¼¬ ¼uv , or, more quickly, 13 1 6 52221 2 2 4−− − + ªº ªºª ºª º− = − == «» «»« »« »− + ¬¼ ¬¼¬ ¼¬ ¼uv . The intermediate step is often not written. 1.3 • Solutions 15 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 2. 32 3 2 5 21 2 1 1+ ªº ª º ª º ªº+= + = =«» « » « » «»−−¬¼ ¬ ¼ ¬ ¼ ¬¼uv . Using the definitions carefully, 32 3 ( 2 ) ( 2 ) 3 4 12( 2 )21 2 ( 2 ) ( 1 ) 2 2 4−− − ªº ª º ªº ª º ª º ª º− =+ − =+ = = «» « » «» « » « » « »−− − + ¬¼ ¬ ¼ ¬¼ ¬ ¼ ¬ ¼ ¬ ¼uv , or, more quickly, 32 3 4 12221 2 2 4−− ªº ª º ª º ª º− = − == «» « » « » « »− + ¬¼ ¬ ¼ ¬ ¼ ¬ ¼uv . The intermediate step is often not written. 3. 4. 5. 1235 2 20 3 89 8xxªº ªºªº «» «»«»−+= −«» «»«» «» «»«» − ¬¼ ¬¼¬¼, 12 1 1235 2 20 3 89 8xx xxxªº ªº ª º «» «» « »− += −«» «» « » «» «» « » − ¬¼ ¬¼ ¬ ¼, 12 1 1235 2 23 89 8xx x xx+ª ºªº « »«»− =−« »«» « »«»−¬ ¼¬¼ 12 1 1235 2 23 89 8xx x xx+= − =− − = Usually the intermediate steps are not displayed. 6. 12 337 2 0 23 1 0xx x− ªº ª º ªºª º++ =«» « » «»« »−¬¼ ¬ ¼ ¬¼¬ ¼, 3 12 3 122 37 0 23 0x xx x xx−ªº ªº ª º ªº++ = «» «» « » «»− ¬¼ ¬¼ ¬ ¼ ¬¼, 123 12 337 2 0 23 0xxx xx x+ − ªº ªº= «» «»− ++ ¬¼ ¬¼ 223 1233720 23 0xxx xxx+ − = − ++= Usually the intermediate steps are not displayed. 7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every vector in R2 can be written as a linear combination of u and v. To write a vector a as a linear combination of u and v, imagine walking from the origin to a along the grid "streets" and keep track of how many "blocks" you travel in the u-direction and how many in the v-direction. a. To reach a from the origin, you might travel 1 unit in the u-direction and –2 units in the v- direction (that is, 2 units in the negative v-direction). Hence a = u – 2v . b. To reach b from the origin, travel 2 units in the u-direction and –2 units in the v-direction. So b = 2u – 2v. Or, use the fact that b is 1 unit in the u-direction from a, so that 16 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. b = a + u = ( u – 2v ) + u = 2u – 2v c. The vector c is –1.5 units from b in the v-direction, so c = b – 1.5v = (2 u – 2v ) – 1.5v = 2u – 3.5v d. The “map” suggests that you can reach d if you travel 3 units in the u-direction and –4 units in the v-direction. If you prefer to stay on the paths displayed on the map, you might travel from the origin to –3 v, then move 3 units in the u-direction, and finally move –1 unit in the v-direction. So d = –3v + 3u – v = 3u – 4v A n o t h e r s o l u t i o n i s d = b – 2v + u = (2 u – 2v ) – 2v + u = 3u – 4v Figure for Exercises 7 and 8 8. See the figure above. Since the grid can be extended in every direction, the figure suggests that every vector in R2 can be written as a linear combination of u and v. w. To reach w from the origin, travel –1 units in the u-direction (that is, 1 unit in the negative u-direction) and travel 2 units in the v-direction. Thus, w = (–1) u + 2v, or w = 2v – u . x. To reach x from the origin, travel 2 units in the v-direction and –2 units in the u-direction. Thus, x = –2u + 2v. Or, use the fact that x is –1 units in the u-direction from w, so that x = w – u = (– u + 2v) – u = –2u + 2v y. The vector y is 1.5 units from x in the v-direction, so y = x + 1.5v = (–2 u + 2v) + 1.5 v = –2u + 3.5v z. The map suggests that you can reach z if you travel 4 units in the v-direction and –3 units in the u-direction. So z = 4v – 3u = –3 u + 4v. If you prefer to stay on the paths displayed on the “map,” you might travel from the origin to –2 u, then 4 units in the v-direction, and finally move –1 unit in the u-direction. So z = –2u + 4v – u = –3 u + 4v 9. 23 123 12350 46 0 380xx xxx xxx+= + − = − + − =, 23 12 3 12350 46 0 38 0xx xx x xxx+ªº ªº «» «»+ − =«» «» «» «» −+ −¬¼ ¬¼ 23 12 3 12 305 0 46 0 38 0xx xx x xx xªº ªº ª º ª º «» «» « » « »++ − =«» «» « » « » «» «» « » « »−−¬¼ ¬¼ ¬ ¼ ¬ ¼, 12 301 5 0 46 1 0 13 8 0xx xªºª ºª º ª º «»« »« » « »++ −=«»« »« » « » «»« »« » « »−−¬¼¬ ¼¬ ¼ ¬ ¼ Usually, the intermediate calculations are not displayed. w xvu acd 2vb zy –2v–u–v01.3 • Solutions 17 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. Note : T h e Study Guide s a y s , “ C h e c k w i t h y o u r i n s t r u c t o r whether you need to “show work” on a problem such as Exercise 9.” 10. 123 123 12332 4 3 27 5 1 54 32xxx xxx xxx− += −− += + − = , 123 123 12332 4 3 27 5 1 543 2xx x xxx xxx− + ªº ªº «» «»−− +=«» «» «» «» + − ¬¼ ¬¼ 123 123 123324 3 275 1 543 2xxx xxxxxx−ªº ªº ªº ª º «» «» «» « »− +− +=«» «» «» « » «» «» «» « » − ¬¼ ¬¼ ¬ ¼ ¬¼ , 123324 3 275 1 543 2xx x− ªºª ºª º ª º «»« »« » « »−+ −+=«»« »« » « » «»« »« » « » − ¬¼¬ ¼¬ ¼ ¬ ¼ Usually, the intermediate calculations are not displayed. 11. The question I s b a linear combination of a1, a2, and a 3? is equivalent to the question Does the vector equation x1a1 + x 2a2 + x 3a3 = b have a solution? T h e e q u a t i o n 12 3 12310 5 2 21 6 1 02 8 6xx xªº ª º ªºªº «» « » «»«»−++ −=−«» « » «»«» «» « » «»«»¬¼ ¬ ¼ ¬¼¬¼ ↑↑↑ ↑ aa a b ( *) has the same solution set as the linear system whose augmented matrix is 10 5 2 21 6 1 02 8 6Mªº «»=−− −«» «»¬¼ R o w r e d u c e M until the pivot positions are visible: 1052 1052 ~0 1 4 3~0 1 4 3 0286 0000Mªº ªº «» «» «» «» «» «»¬¼ ¬¼ The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and therefore b is a linear combination of a1, a2, and a 3. 18 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 12. The equation 12 3 12 3126 1 1 037 5 125 9xx x−− ªº ª º ª º ª º «» « » « » « »++= −«» « » « » « »«» « » « » « » − ¬¼ ¬ ¼ ¬ ¼ ¬ ¼ ↑↑ ↑ ↑ aa a b ( *) has the same solution set as the linear system whose augmented matrix is 126 1 1 0375 1259M−−ªº «»= −«» «»−¬¼ R o w r e d u c e M until the pivot positions are visible: 126 1 1 ~0 3 7 5 00 1 12M−−ªº «»−«» «» − ¬¼ The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and therefore b is a linear combination of a1, a2, and a 3. 13. Denote the columns of A by a 1, a2, a3. To determine if b is a linear combination of these columns, use the boxed fact in the subsection Linear Combinations . Row reduce the augmented matrix [a1 a 2 a 3 b] until you reach echelon form: [ a1 a 2 a 3 b] =1423 14 23 0357 ~ 03 57 2843 00 03−−ª ºª º « »« »−−« »« » « »« » −− −¬ ¼¬ ¼ The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns of A. 14. Row reduce the augmented matrix [ a1 a 2 a 3 b] until you reach echelon form: [a1 a 2 a 3 b] = 10 5 2 1052 1052 21 6 1 ~ 01 43 ~ 01 43 02 8 6 0286 0000ªº ª º ª º «» « » « »−− −«» « » « » «» « » « »¬¼ ¬ ¼ ¬ ¼. The linear system corresponding to this matrix has a solution, so b is a linear combination of the columns of A. 15. [ a1 a 2 b] = 153 15 3 15 3 15 3 385 ~ 07 1 4 ~ 01 2 ~ 01 2 12 03 3 03 3 00 3 hh h h−− − −ªº ª º ª º ª º «» « » « » « »−− − − −«» « » « » « » «» « » « » « »−− + − + − ¬¼ ¬ ¼ ¬ ¼ ¬ ¼. The vector b is in Span{ a1, a2} when h – 3 is zero, that is, when h = 3. 1.3 • Solutions 19 Copyright © 2012 Pearson Education, Inc. Publishing as Addison-Wesley. 16. [ v1 v 2 y] = 12 12 12 013 ~ 01 3~ 01 3 275 035 2 00 4 2hh h hh−− −ªº ª º ª º «» « » « »−− −«» « » « » «» « » « »−− − ++ ¬¼ ¬ ¼ ¬ ¼. The vector y is in Span{ v1, v2} when 4 + 2 h is zero, that is, when h = –2. 17. Noninteger weights are acceptable, of course, but some simple choices are 0 v1 + 0 v2 = 0, and 1 v1 + 0 v2 = 3 1 2ªº «»«» «»¬¼, 0 v1 + 1 v2 = 4 0 1−ªº «» «» «»¬¼, 1 v1 + 1 v2 = 1 1 3−ªº «» «» «»¬¼, 1 v1 – 1 v2 = 7 1 1ªº «»«» «»¬¼ 18. Some likely choices are 0 v1 + 0 v2 = 0, and 1 v1 + 0 v2 = 1 1 2ªº «»«» «»−¬¼, 0 v1 + 1 v2 = 2 3 0−ªº «»«» «»¬¼, 1 v1 + 1 v2 = 1 4 2−ªº «» «» «»−¬¼, 1 v1 – 1 v2 = 3 2 2ªº «»−«» «»−¬¼ 19. By inspection, v2 = (3/2) v1. Any linear combination of v1 and v 2 is actually just a multiple of v1. For instance, a v1 + bv 2 = av1 + b(3/2) v1 = (a + 3 b/2)v 1 S o S p a n { v1, v2} is the set of points on the line through v1 and 0. Note : Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7. 20. Span{ v1, v2} is a plane in R3 through the origin, because neither vector in this problem is a multiple of the other. 21. Let y = h kªº «»¬¼. Then [ u v y] = 22 22~11 02 / 2hh kk hªº ª º «» « »− + ¬¼ ¬ ¼. This augmented matrix corresponds to a consistent system for all h and k. So y is in Span{ u, v} for all h and k . 22. Construct any 3 ×4 matrix in echelon form that corresponds to an inconsistent system. Perform sufficient row operations on the matrix to eliminate all zero entries in the first three columns. 23. a. False. The alternative notation for a (column) vector is (–4, 3), using parentheses and commas. b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If 5 2−ªº «»¬¼ were on the line through 2 5−ªº «»¬¼ and the origin, then 5 2−ªº «»¬¼ would have to be a multiple of 2 5−ªº «»¬¼, which is not the case. c. True. See the line displayed just before Example 4. d. True. See the box that discusses the matrix in (5). 20 CHAPTER 1 • Linear Equations in Linear Algebra Copyright ©! 2012 Pearson Education, Inc. Publishing as Addison-Wesley. e. False. The statement is often true, but Span{ u, v} is not a plane when v is a multiple of u, or when u is the zero vector. 24. a. False. Span{ u, v} can be a plane. b. True. See the beginning of the subsection Vectors in Rn. c. True. See the comment following the definition of Span{ v1, …, v p}. d. False. ( u – v) + v = u – v + v = u. e. False. Setting all the weights equal to zero results in a legitimate linear combination of a set of vectors. 25. a. There are only three vectors in the set {a 1, a2, a3}, and b is not one of them. b. There are infinitely many vectors in W = Span{ a1, a2, a3}. To determine if b is in W, use the method of Exercise 13. [a1 a 2 a 3 b] =10 4 4 10 44 10 44 03 2 1 ~ 03 21 ~ 03 21 26 3 4 06 54 00 12−− − ª ºª ºª º « »« »« »−− −« »« »« » « »« »« » −− − −¬ ¼¬ ¼¬ ¼ The system for this augmented matrix is consistent, so b is in W. c. a1 = 1a 1 + 0a 2 + 0a3. See the discussion in the text following the definition of Span{ v1, …, v p}. 26. a. [a1 a 2 a 3 b] = 20 6 1 0 10 3 5 103 5 1 0 3 5 18 53 ~ 18 5 3 ~ 088 8 ~ 0 8 8 8 12 1 7 12 1 7 022 2 0 0 0 4−− −− − −ª ºª ºª ºª º « »« »« »« » « »« »« »« » « »« »« »« » ¬ ¼¬ ¼¬ ¼¬ ¼ N o , b is not a linear combination of the columns of A, that is, b is not in W. b. The second column of A is in W because a2 = 0 a1 + 1 a2 + 0 a3. 27. a. 5v 1 is the output of 5 days’ operation of mine #1. b. The total output is x1v1 + x 2v2, so x 1 and x 2 should satisfy 11 2 2240 2824xxªº+=«»¬¼vv . c. [M] Reduce the augmented matrix 30 40 240 1 0 1.73~600 380 2824 0 1 4.70ª ºª º « »« »¬ ¼¬ ¼. Operate mine #1 for 1.73 days and mine #2 for 4.70 days. (This is an approximate solution.) 28. a. The amount of heat produced when the steam plant burns x1 tons of anthracite and x2 tons of bituminous coal is 27.6 x1 + 30.2x 2 million Btu. b. The total output produced by x1 tons of anthracite and x2 tons of bituminous coal is given by the vector 1227.6 30.2 3100 6400 250 360xxªºªº «»«»+«»«» «»«»¬¼¬¼. c. [M] The appropriate values for x1 and x2 satisfy 1227.6 30.2 162 3100 6400 23,610 250 360 1,623xxªºªº ª º «»«» « »+=«»«» « » «»«» « »¬¼¬¼ ¬ ¼. To solve, row reduce the augmented matrix: