 Copyright © 2016 Pearson Education, Inc. 1-1 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 12 57 275 xx xx 157 275 Replace R2 by R2 + (2)R1 and obtain: 12 2 57 39 xx x 157 039 Scale R2 by 1/3: 12 2 57 3 xx x 157 013 Replace R1 by R1 + (–5)R2: 1 2 8 3 x x 108 013 The solution is (x1, x2) = (–8, 3), or simply (–8, 3). 2. 12 12 244 5711 xx xx 244 5711 Scale R1 by 1/2 and obtain: 12 12 22 5711 xx xx 122 5711 Replace R2 by R2 + (–5)R1: 12 2 22 321 xx x 122 0321 Scale R2 by –1/3: 12 2 22 7 xx x 122 017 Replace R1 by R1 + (–2)R2: 1 2 12 7 x x 1012 017 The solution is (x1, x2) = (12, –7), or simply (12, –7).
2. 2 CHAPTER 1 Linear Equations in Linear Algebra3. The point of intersection satisfies the system of two linear equations:x xx x+ = =5 72 21 21 21 5 71 2 2 x x+ = =5 77 9Replace R2 by R2 + (1)R1 and obtain: 1 2x21 5 70 7 9 + =5 7Scale R2 by 1/7: 1 229/7x xx=1 5 70 1 9/7 Replace R1 by R1 + (5)R2: 124/79/7xx==1 0 4/70 1 9/7 The point of intersection is (x1, x2) = (4/7, 9/7).4. The point of intersection satisfies the system of two linear equations:x xx x = =5 11 21 23 7 5 1 5 1 3 7 5 x x =5 18 2Replace R2 by R2 + (3)R1 and obtain: 1 2x2= 1 5 1 0 8 2 =5 1Scale R2 by 1/8: 1 221/4x xx= 1 5 1 0 1 1/4Replace R1 by R1 + (5)R2: 129/41/4xx==1 0 9/40 1 1/4 The point of intersection is (x1, x2) = (9/4, 1/4).5. The system is already in triangular form. The fourth equation is x4 = 5, and the other equations do notcontain the variable x4. The next two steps should be to use the variable x3 in the third equation toeliminate that variable from the first two equations. In matrix notation, that means to replace R2 by itssum with 3 times R3, and then replace R1 by its sum with 5 times R3.6. One more step will put the system in triangular form. Replace R4 by its sum with 3 times R3, whichproduces 1 6 4 0 1 0 2 7 0 4 0 0 1 2 3 0 0 0 5 15 . After that, the next step is to scale the fourth row by 1/5.7. 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 x3 = 1,or simply, 0 = 1. A system containing this condition has no solution. Further row operations areunnecessary once an equation such as 0 = 1 is evident.The solution set is empty. 3. 1.1 Solutions 38. The standard row operations are: 1 4 9 0 1 4 9 0 1 4 0 0 1 0 0 0 0 1 7 0 ~ 0 1 7 0 ~ 0 1 0 0 ~ 0 1 0 0 0 0 2 0 0 0 1 0 0 0 1 0 0 0 1 0The solution set contains one solution: (0, 0, 0).9. The system has already been reduced to triangular form. Begin by scaling the fourth row by 1/2 and thenreplacing R3 by R3 + (3)R4: 1 1 0 0 4 1 1 0 0 4 1 1 0 0 4 0 1 3 0 7 ~ 0 1 3 0 7 ~ 0 1 3 0 7 0 0 1 3 1 0 0 1 3 1 0 0 1 0 5 0 0 0 2 4 0 0 0 1 2 0 0 0 1 2Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2: 1 1 0 0 4 1 0 0 0 40 1 0 0 8 0 1 0 0 8~ ~0 0 1 0 5 0 0 1 0 50 0 0 1 2 0 0 0 1 2The solution set contains one solution: (4, 8, 5, 2).10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the4 and 3 above it to zeros. That is, replace R2 by R2 + (4)R4 and replace R1 by R1 + (3)R4. For thefinal step, replace R1 by R1 + (2)R2. 1 2 0 3 2 1 2 0 0 7 1 0 0 0 3 0 1 0 4 7 0 1 0 0 5 ~ 0 1 0 0 5~ 0 0 1 0 6 0 0 1 0 6 0 0 1 0 6 0 0 0 1 3 0 0 0 1 3 0 0 0 1 3The solution set contains one solution: (3, 5, 6, 3).11. First, swap R1 and R2. Then replace R3 by R3 + (3)R1. Finally, replace R3 by R3 + (2)R2. 0 1 4 5 1 3 5 2 1 3 5 2 1 3 5 2 1 3 5 2 ~ 0 1 4 5 ~ 0 1 4 5 ~ 0 1 4 5 3 7 7 6 3 7 7 6 0 2 8 12 0 0 0 2The system is inconsistent, because the last row would require that 0 = 2 if there were a solution.The solution set is empty.12. Replace R2 by R2 + (3)R1 and replace R3 by R3 + (4)R1. Finally, replace R3 by R3 + (3)R2. 1 3 4 4 1 3 4 4 1 3 4 4 3 7 7 8 ~ 0 2 5 4 ~ 0 2 5 4 4 6 1 7 0 6 15 9 0 0 0 3The system is inconsistent, because the last row would require that 0 = 3 if there were a solution.The solution set is empty. 4. 4 CHAPTER 1 Linear Equations in Linear Algebra13. 1 0 3 8 1 0 3 8 1 0 3 8 1 0 3 8 2 2 9 7 ~ 0 2 15 9 ~ 0 1 5 2 ~ 0 1 5 2 0 1 5 2 0 1 5 2 0 2 15 9 0 0 5 51 0 3 8 1 0 0 5~ 0 1 5 2 ~ 0 1 0 30 0 1 1 0 0 1 1 . The solution is (5, 3, 1).14. 1 3 0 5 1 3 0 5 1 3 0 5 1 3 0 5 1 1 5 2 ~ 0 2 5 7 ~ 0 1 1 0 ~ 0 1 1 0 0 1 1 0 0 1 1 0 0 2 5 7 0 0 7 71 3 0 5 1 3 0 5 1 0 0 2~ 0 1 1 0 ~ 0 1 0 1 ~ 0 1 0 1 .0 0 1 1 0 0 1 1 0 0 1 1 The solution is (2, 1, 1).15. First, replace R4 by R4 + (3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 by R4 + (3)R3.1 0 3 0 2 1 0 3 0 20 1 0 3 3 0 1 0 3 3 ~ 0 2 3 2 1 0 2 3 2 1 3 0 0 7 5 0 0 9 7 11 1 0 3 0 2 1 0 3 0 20 1 0 3 3 0 1 0 3 3~ ~0 0 3 4 7 0 0 3 4 70 0 9 7 11 0 0 0 5 10 The resulting triangular system indicates that a solution exists. In fact, using the argument from Example 2,one can see that the solution is unique.16. First replace R4 by R4 + (2)R1 and replace R4 by R4 + (3/2)R2. (One could also scale R2 beforeadding to R4, but the arithmetic is rather easy keeping R2 unchanged.) Finally, replace R4 by R4 + R3. 1 0 0 2 3 1 0 0 2 3 0 2 2 0 0 0 2 2 0 0~ 0 0 1 3 1 0 0 1 3 1 2 3 2 1 5 0 3 2 3 1 1 0 0 2 3 1 0 0 2 30 2 2 0 0 0 2 2 0 0~ ~0 0 1 3 1 0 0 1 3 10 0 1 3 1 0 0 0 0 0 The system is now in triangular form and has a solution. The next section discusses how to continue withthis type of system. 5. 1.1 Solutions 517. Row reduce the augmented matrix corresponding to the given system of three equations: 1 4 1 1 4 1 1 4 1 2 1 3 ~ 0 7 5 ~ 0 7 5 1 3 4 0 7 5 0 0 0The system is consistent, and using the argument from Example 2, there is only one solution. So the threelines have only one point in common.18. Row reduce the augmented matrix corresponding to the given system of three equations:1 2 1 4 1 2 1 4 1 2 1 40 1 1 1 ~ 0 1 1 1 ~ 0 1 1 11 3 0 0 0 1 1 4 0 0 0 5 The third equation, 0 = 5, shows that the system is inconsistent, so the three planes have no point incommon.19.h h2 4 1 4 ~ 3 6 8 0 6 3 h 4Write c for 6 3h. If c = 0, that is, if h = 2, then the system has nosolution, because 0 cannot equal 4. Otherwise, when h 2, the system has a solution.20.h h 1 3 1 3 + ~ .h2 4 6 0 4 2 0Write c for 4 + 2h. Then the second equation cx2 = 0 has a solutionfor every value of c. So the system is consistent for all h.21. 1 3 2 1 3 2 + ~ .4 h 8 0 h 12 0Write c for h + 12. Then the second equation cx2 = 0 has a solutionfor every value of c. So the system is consistent for all h.22.h h 2 3 2 3 + ~ .h6 9 5 0 0 5 3The system is consistent if and only if 5 + 3h = 0, that is, if and onlyif h = 5/3.23. 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 given applied to a single solution. The solution set consists of all possiblesolutions. Only in special cases does the solution set consist of exactly one solution. Mark a statementTrue only if the statement is always true.d. True. See the box before Example 2.24. a. True. See the box preceding the subsection titled Existence and Uniqueness Questions.b. False. The definition of row equivalent requires that there exist a sequence of row operations thattransforms one matrix into the other.c. False. By definition, an inconsistent system has no solution.d. True. This definition of equivalent systems is in the second paragraph after equation (2). 6. 6 CHAPTER 1 Linear Equations in Linear Algebra25.g g gh h hk k g k g h 1 4 7 1 4 7 1 4 7 0 3 5 ~ 0 3 5 ~ 0 3 5 2 5 9 0 3 5 + 2 0 0 0 + 2+ Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix aboveis 0 = b. This equation is possible if and only if b is zero. So the original system has a solution if and onlyif k + 2g + h = 0.26. 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 (2, 1, 0), and then performany elementary row operations to produce other augmented matrices. Here are three examples. The factthat they are all row equivalent proves that they all have the solution set (2, 1, 0). 1 0 0 2 1 0 0 2 1 0 0 2 0 1 0 1 ~ 2 1 0 3 ~ 2 1 0 3 0 0 1 0 0 0 1 0 2 0 1 427. Study the augmented matrix for the given system, replacing R2 by R2 + (c)R1:f f1 3 1 3 ~ c d g 0 d 3c g cfThis shows that shows d 3c must be nonzero, since f and g are arbitrary. Otherwise, for some choicesof f and g the second row would correspond to an equation of the form 0 = b, where b is nonzero.Thus d 3c.28. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possiblesince a is nonzero. Then replace R2 by R2 + (c)R1.a b f 1 b / a f / a 1 b / a f /a~ ~c d g c d g d cb a g c f a 0 (/) (/) The quantity d c(b/a) must be nonzero, in order for the system to be consistent when the quantityg c( f /a) is nonzero (which can certainly happen). The condition that d c(b/a) 0 can also be writtenas ad bc 0, or ad bc.29. Swap R1 and R2; swap R1 and R2.30. Multiply R2 by 1/2; multiply R2 by 2.31. Replace R3 by R3 + (4)R1; replace R3 by R3 + (4)R1.32. Replace R3 by R3 + (3)R2; replace R3 by R3 + (3)R2.33. The first equation was given. The others are:T2 = (T1 + 20 + 40 + T3 )/4, or 4T2 T1 T3 = 60T3 = (T4 + T2 + 40 + 30)/4, or 4T3 T4 T2 = 70T4 = (10 + T1 + T3 + 30)/4, or 4T4 T1 T3 = 40 7. 1.1 Solutions 7Rearranging,T T TT T T =4 301 2 41 2 3 + =4 60T T T + =4 702 3 4T T T + =4 401 3 434. Begin by interchanging R1 and R4, then create zeros in the first column: 4 1 0 1 30 1 0 1 4 40 1 0 1 4 40 1 4 1 0 60 ~ 1 4 1 0 60 0 4 0 4 20~ 0 1 4 1 70 0 1 4 1 70 0 1 4 1 70 1 0 1 4 40 4 1 0 1 30 0 1 4 15 190 Scale R1 by 1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3: 1 0 1 4 40 1 0 1 4 40 1 0 1 4 400 1 0 1 5 0 1 0 1 5 0 1 0 1 5~ ~ ~0 1 4 1 70 0 0 4 2 75 0 0 4 2 750 1 4 15 190 0 0 4 14 195 0 0 0 12 270Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 1 0 1 4 40 1 0 1 0 50 1 0 1 0 500 1 0 1 5 0 1 0 0 27.5 0 1 0 0 27.5~ ~ ~0 0 4 2 75 0 0 4 0 120 0 0 1 0 300 0 0 1 22.5 0 0 0 1 22.5 0 0 0 1 22.5The last step is to replace R1 by R1 + (1)R3:1 0 0 0 20.00 1 0 0 27.5~ .0 0 1 0 30.00 0 0 1 22.5 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, eventhe small grid for Exercise 34 leads to about 25 multiplications or additions (not counting operations withzero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14in Section 1.10 returns to this problem and states the solution in case students have not already solved thesystem of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LUfactorization.For instructors who wish to use technology in the course, the Study Guide provides boxed MATLABnotes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/86/89 and HP-48Gcalculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section 1.1describes how to access the data that is available for all numerical exercises in the text. This feature has theability 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 areincluded in the text data sets, available from the text web site, www.laylinalgebra.com. 8. 8 CHAPTER 1 Linear Equations in Linear Algebra1.2 SOLUTIONSNotes: The key exercises are 120 and 2328. (Students should work at least four or five from Exercises714, 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.3.1 2 3 4 1 2 3 4 1 2 3 44 5 6 7 ~ 0 3 6 9 ~ 0 1 2 36 7 8 9 0 5 10 15 0 5 10 15 1 2 3 4 1 0 1 2~ 0 1 2 3 ~ 0 1 2 30 0 0 0 0 0 0 0. Pivot cols 1 and 2.1 2 3 44 5 6 76 7 8 9 4.1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 73 5 7 9 ~ 0 4 8 12 ~ 0 1 2 3 ~ 0 1 2 35 7 9 1 0 8 16 34 0 8 16 34 0 0 0 10 1 3 5 7 1 3 5 0 1 0 1 0~ 0 1 2 3 ~ 0 1 2 0 ~ 0 1 2 00 0 0 1 0 0 0 1 0 0 0 1. Pivot cols1, 2, and 41 3 5 73 5 7 95 7 9 1 5.* * 0, , 0 0 0 0 06.* * 0 0 ,0 0,0 0 0 0 0 0 0 07. 1 3 4 7 1 3 4 7 1 3 4 7 1 3 0 5 ~ ~ ~3 9 7 6 0 0 5 15 0 0 1 3 0 0 1 3 1 + 3 Corresponding system of equations: 2= 533x xx=The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is free.Solve for the basic variables in terms of the free variable. The general solution isx xxx= =5 3is free31 223Note: Exercise 7 is paired with Exercise 10. 9. 1.2 Solutions 98. 1 4 0 7 1 4 0 7 1 4 0 7 1 0 0 9 ~ ~ ~ 2 7 0 10 0 1 0 4 0 1 0 4 0 1 0 4 Corresponding system of equations: 1294xx= =The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is free.Solve for the basic variables in terms of the free variable. In this particular problem, the basic variablesdo not depend on the value of the free variable.General solution:12394is freexxx= = Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic variablesfirst. Any remaining variables are free. (This type of computation will arise in Chapter 5.)9. 0 1 6 5 1 2 7 6 1 0 5 4 1 2 7 6 ~ ~ 0 1 6 5 0 1 6 5 x 1 xCorresponding system: 3 = =5 46 5x x2 3Basic variables: x1, x2; free variable: x3. General solution:x xx xx= + 4 55 6is free1 32 33= + 10. 1 2 1 3 1 2 1 3 1 2 0 4 ~ ~ 3 6 2 2 0 0 1 7 0 0 1 7 1 2 Corresponding system: 2= 437x xx= Basic variables: x1, x3; free variable: x2. General solution:x xxx= + = 4 2is free71 22311. 3 4 2 0 3 4 2 0 1 4/3 2/3 0 9 12 6 0 ~ 0 0 0 0 ~ 0 0 0 0 6 8 4 0 0 0 0 0 0 0 0 0Corresponding system:4 2 03 3x x + x =1 2 3==0 00 0 10. 10 CHAPTER 1 Linear Equations in Linear AlgebraBasic variable: x1; free variables x2, x3. General solution: = 1 2 3234 23 3is freeis freex x xxx12. 1 7 0 6 5 1 7 0 6 5 1 7 0 6 5 0 0 1 2 3 ~ 0 0 1 2 3 ~ 0 0 1 2 3 1 7 4 2 7 0 0 4 8 12 0 0 0 0 0Corresponding system:x x x + =7 6 51 2 4x x =2 30 03 4=Basic variables: x1 and x3; free variables: x2, x4. General solution:x x xxx xx= + 5 7 6is free3 2is free1 2 423 44= + 13. 1 3 0 1 0 2 1 3 0 0 9 2 1 0 0 0 3 5 0 1 0 0 4 1 0 1 0 0 4 1 0 1 0 0 4 1 ~ ~ 0 0 0 1 9 4 0 0 0 1 9 4 0 0 0 1 9 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0Corresponding system:x x = =+ =3 54 19 40 01 5x x2 5x x4 5=Basic variables: x1, x2, x4; free variables: x3, x5. General solution:x xx xxx xx= + 5 31 4is free4 9is free1 52 534 55= + = Note: The Study Guide discusses the common mistake x3 = 0.14. 1 2 5 6 0 5 1 0 7 0 0 9 0 1 6 3 0 2 0 1 6 3 0 2 ~ 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 11. 1.2 Solutions 11Corresponding system:+ = =7 96 3 21 32 3 4500 0x xx x xx==Basic variables: x1, x2, x5; free variables: x3, x4. General solution:x xx x xxxx= 9 72 6 3is freeis free01 32 3 4345= + + = 15. a. The system is consistent, with a unique solution.b. The system is inconsistent. (The rightmost column of the augmented matrix is a pivot column).16. a. The system is consistent, with a unique solution.b. The system is consistent. There are many solutions because x2 is a free variable.17.h h2 3 2 3 ~ 4 6 7 0 0 7 2hThe system has a solution only if 7 2h = 0, that is, if h = 7/2.18. 1 3 2 1 3 2 ~ 5 h 7 0 h + 15 3If h +15 is zero, that is, if h = 15, then the system has no solution,because 0 cannot equal 3. Otherwise, when h 15, the system has a solution.19.h h2 2 1 2 ~ 4 8 k 0 8 4 h k 8 a. When h = 2 and k 8, the augmented column is a pivot column, and the system is inconsistent.b. When h 2, 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.20.1 3 2 1 3 2 ~ 3 h k 0 h 9 k 6 a. When h = 9 and k 6, the system is inconsistent, because the augmented column is a pivot column.b. When h 9, the system is consistent and has a unique solution. There are no free variables.c. When h = 9 and k = 6, 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 5x4 = 0, which does not by itself lead to acontradiction. So the system might be consistent or it might be inconsistent. 12. 12 CHAPTER 1 Linear Equations in Linear Algebra22. a. False. See the statement preceding Theorem 1. Only the reduced echelon form is unique.b. False. See the beginning of the subsection Pivot Positions. The pivot positions in a matrix aredetermined completely by the positions of the leading entries in the nonzero rows of any echelonform obtained from the matrix.c. True. See the paragraph after Example 3.d. 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.e. True. See the paragraph just before Example 4.23. Yes. The system is consistent because with three pivots, there must be a pivot in the third (bottom) rowof the coefficient matrix. The reduced echelon form cannot contain a row of the form[0 0 0 0 0 1].24. The system is inconsistent because the pivot in column 5 means that there is 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 systemhas no solution.25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottomrow, and there is no room for a pivot in the augmented column. So, the system is consistent, byTheorem 2.26. Since there are three pivots (one in each row), the augmented matrix must reduce to the forma x ab x bc x c 1 0 0 1= 0 1 0 and so2= 0 0 1 3=No matter what the values of a, b, and c, the solution exists and is unique.27. If a linear system is consistent, then the solution is unique if and only if every column in the coefficientmatrix is a pivot column; otherwise there are infinitely many solutions. This statement is true because the free variables correspond to nonpivot columns of the coefficientmatrix. The columns are all pivot columns if and only if there are no free variables. And there are no freevariables 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 rightmostcolumn is not a pivot column.29. An underdetermined system always has more variables than equations. There cannot be more basicvariables than there are equations, so there must be at least one free variable. Such a variable may beassigned infinitely many different values. If the system is consistent, each different value of a freevariable will produce a different solution.x x xx x x+ + =+ + =30. Example: 1 2 31 2 342 2 2 531. Yes, a system of linear equations with more equations than unknowns can be consistent.Example (in which x1 = x2 = 1):x xx xx x+ = =+ =1 21 21 2203 2 5 13. 1.2 Solutions 1332. According to the numerical note in Section 1.2, when n = 30 the reduction to echelon form takes about2(30)3/3 = 18,000 flops, while further reduction to reduced echelon form needs at most (30)2 = 900 flops.Of the total flops, the backward phase is about 900/18900 = .048 or about 5%.When n = 300, the estimates are 2(300)3/3 = 18,000,000 phase for the reduction to echelon form and(300)2 = 90,000 flops for the backward phase. The fraction associated with the backward phase is about(9104) /(18106) = .005, or about .5%.33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 12), (2, 15), and (3, 16), thecoefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the augmentedmatrix:1 1 1 12 1 1 1 12 1 1 1 12 1 1 1 121 2 4 15 ~ 0 1 3 3 ~ 0 1 3 3 ~ 0 1 3 31 3 9 16 0 2 8 4 0 0 2 2 0 0 1 1 1 1 0 13 1 0 0 7~ 0 1 0 6 ~ 0 1 0 60 0 1 1 0 0 1 1 The polynomial is p(t) = 7 + 6t t2.34. [M] The system of equations to be solved is:2 3 4 5a a a a a aa a a a a aa a a a a aa a a a a aa a a a a aa a a a+ + + + + =+ + + + + =+ + + + + =+ + + + + =+ + + + + =+ + + + 4 50 0 0 0 0 02 2 2 2 2 2.904 4 4 4 4 14.86 6 6 6 6 39.68 8 8 8 8 74.310 10 100 1 2 3 4 52 3 4 50 1 2 3 4 52 3 4 50 1 2 3 4 52 3 4 50 1 2 3 4 52 3 4 50 1 2 3 4 52 30 1 2 3a4 10 + a5 10 = 119The unknowns are a0, a1, , a5. Use technology to compute the reduced echelon of the augmentedmatrix:1 0 0 0 0 0 0 1 0 0 0 0 0 01 2 4 8 16 32 2.9 0 2 4 8 16 32 2.91 4 16 64 256 1024 14.8 0 0 8 48 224 960 9 ~ 2 3 4 5 4.5 1 6 36 216 1296 7776 39.6 0 0 24 192 1248 7680 30.91 8 64 512 4096 32768 74.3 0 0 48 480 4032 32640 62.71 10 10 10 10 10 119 0 0 80 960 9920 99840 101 0 0 0 0 0 0 1 0 0 0 0 0 00 2 4 8 16 32 2.9 0 2 4 8 16 32 2.90 0 8 48 224 960 9 0 0 8 48 224 960 9~ ~0 0 0 48 576 4800 3.9 0 0 0 48 576 4800 3.90 0 0 192 2688 26880 8.7 0 0 0 0 384 7680 6.90 0 0 480 7680 90240 14.5 0 0 0 0 1920 42240 24.5 14. 14 CHAPTER 1 Linear Equations in Linear Algebra1 0 0 0 0 0 0 1 0 0 0 0 0 00 2 4 8 16 32 2.9 0 2 4 8 16 32 2.90 0 8 48 224 960 9 0 0 8 48 224 960 9~ ~0 0 0 48 576 4800 3.9 0 0 0 48 576 4800 3.90 0 0 0 384 7680 6.9 0 0 0 0 384 7680 6.90 0 0 0 0 3840 10 0 0 0 0 0 1 .0026 1 0 0 0 0 0 0 1 0 0 0 0 0 00 2 4 8 16 0 2.8167 0 1 0 0 0 0 1.71250 0 8 48 224 0 6.5000 0 0 1 0 0 0 1.1948~ ~~0 0 0 48 576 0 8.6000 0 0 0 1 0 0 .66150 0 0 0 384 0 26.900 0 0 0 0 1 0 .07010 0 0 0 0 1 .002604 0 0 0 0 0 1 .0026 " Thus p(t) = 1.7125t 1.1948t2 + .6615t3 .0701t4 + .0026t5, 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 augmentedmatrix for such a system is the same as the one used to find p, except that at least column 6 is missing. Whenthe augmented matrix is row reduced, the sixth row of the augmented matrix will be entirely zero except for anonzero entry in the augmented column, indicating that no solution exists.Exercise 34 requires 25 row operations. It should give students an appreciation for higher-levelcommands 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 SOLUTIONSNotes: The key exercises are 1114, 1722, 25, and 26. A discussion of Exercise 25 will help studentsunderstand the notation [a1 a2 a3], {a1, a2, a3}, and Span{a1, a2, a3}.1. + 1 3 1 (3) 42 1 2 (1) 1u + v = + = = + . Using the definitions carefully, + 1 3 1 ( 2)( 3) 1 6 5u v , or, more quickly, 2 = + ( 2)2 1 = + = = 2 ( 2)( 1) 2 + 2 4 + 1 3 16 5u v . The intermediate step is often not written. 2 = 2 = = 2 1 2 + 2 42. + 3 2 3 2 52 1 2 (1) 1u + v = + = = + . Using the definitions carefully, 15. 1.3 Solutions 15 + 3 2 3 ( 2)(2) 3 ( 4) 1u v , or, more quickly, 2 = + ( 2)2 1 = + = = 2 ( 2)( 1) 2 + 2 4 3 2 3 4 1u v . The intermediate step is often not written. 2 = 2 = = 2 1 2 + 2 43.x2x1u 2v 2vu v vvuu + v4.x2x1u vuvu + v v 2vu 2v 6 3 1 1 + 4 = 7 5 0 55. x 1 x2,x xx xx 6 1 3 2 1 1 + 4 2 = 7 5 1 0 5,x xx xx 6 1 3 2 1 + 4 1 2 = 7 5 1 5x xx xx =6 3 11 21 21 + = 4 7= 5 5Usually the intermediate steps are not displayed. 2 8 1 0 + 3 5 + = 6 06. x 1 x 2 x3x x xx x x 2 8 0 + 3 + = 5 6 0, 1 2 31 2 3x x xx x x 2 + 8 + 0 = 3 + 5 6 0, 1 2 31 2 3x x xx x x + + =2 8 03 5 6 02 2 31 2 3+ =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 everyvector 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 thegrid "streets" and keep track of how many "blocks" you travel in the u-direction and how many in thev-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. 16. 16 CHAPTER 1 Linear Equations in Linear Algebrab. To reach b from the origin, travel 2 units in the u-direction and 2 units in the v-direction. Sob = 2u 2v. Or, use the fact that b is 1 unit in the u-direction from a, so thatb = a + u = (u 2v) + u = 2u 2vc. The vector c is 1.5 units from b in the v-direction, soc = b 1.5v = (2u 2v) 1.5v = 2u 3.5vd. The map suggests that you can reach d if you travel 3 units in the u-direction and 4 units in thev-direction. If you prefer to stay on the paths displayed on the map, you might travel from the originto 3v, then move 3 units in the u-direction, and finally move 1 unit in the v-direction. Sod = 3v + 3u v = 3u 4vAnother solution isd = b 2v + u = (2u 2v) 2v + u = 3u 4vwvxuadc2vbyzv2v u0Figure for Exercises 7 and 88. See the figure above. Since the grid can be extended in every direction, the figure suggests that everyvector 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 negativeu-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 thatx = w u = (u + 2v) u = 2u + 2vy. The vector y is 1.5 units from x in the v-direction, soy = x + 1.5v = (2u + 2v) + 1.5v = 2u + 3.5vz. The map suggests that you can reach z if you travel 4 units in the v-direction and 3 units in theu-direction. So z = 4v 3u = 3u + 4v. If you prefer to stay on the paths displayed on the map, youmight travel from the origin to 2u, then 4 units in the v-direction, and finally move 1 unit inthe u-direction. Soz = 2u + 4v u = 3u + 4v9.x x+ =5 02 3x x xx x x+ =4 6 01 2 31 2 3 + =3 8 0,x xx x xx x x + 5 2 3 0 4 1 + 6 2 3 = 0 + 3 8 1 2 3 0x x0 5 04 6 0 2 3 x 1 + x 2 + x3 = x 1 3 x 2 8 x3 00 1 5 04 6 1 01 3 8 0 + + = x x x, 1 2 3Usually, the intermediate calculations are not displayed. 17. 1.3 Solutions 17Note: The Study Guide says, Check with your instructor whether you need to show work on a problemsuch as Exercise 9.10.x x xx x xx x x+ + = =+ =4 3 91 2 31 2 31 2 37 2 28 6 5 15,x x xx x xx x x 4 1 + 2 + 3 3 9 1 7 2 2 3 = 2 8 + 6 5 1 2 3 15x x xx x xx x x4 3 9 1 2 3 + 7 1 2 + 2 3 = 2 8 1 6 2 5 3 154 1 3 91 7 2 28 6 5 15 + + = x x x, 1 2 3Usually, the intermediate calculations are not displayed.11. The questionIs b a linear combination of a1, a2, and a3?is equivalent to the questionDoes the vector equation x1a1 + x2a2 + x3a3 = b have a solution?The equation1 0 5 22 1 6 10 2 8 6 + + = a a a bx x x1 2 31 2 3(*)has the same solution set as the linear system whose augmented matrix is1 0 5 22 1 6 10 2 8 6M = Row reduce M until the pivot positions are visible:1 0 5 2 1 0 5 2~ 0 1 4 3 ~ 0 1 4 30 2 8 6 0 0 0 0M The linear system corresponding to M has a solution, so the vector equation (*) has a solution, andtherefore b is a linear combination of a1, a2, and a3.12. The equation 1 0 2 5 2 + 5 + 0 = 11 2 5 8 7 a a a bx x x1 2 31 2 3(*)has the same solution set as the linear system whose augmented matrix is 18. 18 CHAPTER 1 Linear Equations in Linear Algebra1 0 2 52 5 0 112 5 8 7M = Row reduce M until the pivot positions are visible:1 0 2 5 1 0 2 5~ 0 5 4 1 ~ 0 5 4 10 5 4 3 0 0 0 2M The linear system corresponding to M has no solution, so the vector equation (*) has no solution, andtherefore b is not a linear combination of a1, a2, and a3.13. Denote the columns of A by a1, a2, a3. To determine if b is a linear combination of these columns, use theboxed fact on page 34. Row reduced the augmented matrix until you reach echelon form: 1 4 2 3 1 4 2 3 0 3 5 7 ~ 0 3 5 7 2 8 4 3 0 0 0 3The system for this augmented matrix is inconsistent, so b is not a linear combination of the columnsof A.14. [a1 a2 a3 b] = 1 2 6 11 1 2 6 11 0 3 7 5 ~ 0 3 7 5 1 2 5 9 0 0 11 2. The linear system corresponding to thismatrix has a solution, so b is a linear combination of the columns of A.15. Noninteger weights are acceptable, of course, but some simple choices are 0v1 + 0v2 = 0, and1v1 + 0v2 =716 , 0v1 + 1v2 = 5 3 01v1 + 1v2 =246 , 1v1 1v2 =1226 16. Some likely choices are 0v1 + 0v2 = 0, and1v1 + 0v2 =302 , 0v1 + 1v2 = 2 0 31v1 + 1v2 =105 , 1v1 1v2 =501 19. 1.3 Solutions 1917. [a1 a2 b] = 1 2 4 1 2 4 1 2 4 1 2 4 4 3 1 ~ 0 5 15 ~ 0 1 3 ~ 0 1 3 2 7 h 0 3 h + 8 0 3 h + 8 0 0 h + 17. The vector b isin Span{a1, a2} when h + 17 is zero, that is, when h = 17.18. [v1 v2 y] =h h h 1 3 1 3 1 3 0 1 5 ~ 0 1 5 ~ 0 1 5 2 8 3 0 2 3 + 2 h 0 0 7 + 2h. The vector y is inSpan{v1, v2} when 7 + 2h is zero, that is, when h = 7/2.19. By inspection, v2 = (3/2)v1. Any linear combination of v1 and v2 is actually just a multiple of v1. Forinstance,av1 + bv2 = av1 + b(3/2)v2 = (a + 3b/2)v1So Span{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 the neither vector in this problem is a multipleof the other. Every vector in the set has 0 as its second entry and so lies in the xz-plane in ordinary3-space. So Span{v1, v2} is the xz-plane.21. Let y =hk . Then [u v y] =h hk k h2 2 2 2 ~ 1 1 0 2 + /2. This augmented matrix corresponds toa consistent system for all h and k. So y is in Span{u, v} for all h and k.22. Construct any 34 matrix in echelon form that corresponds to an inconsistent system. Perform sufficientrow 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 2were onthe line through 2 5and the origin, then 5 2would have to be a multiple of 2 5, which is notthe case.c. True. See the line displayed just before Example 4.d. True. See the box that discusses the matrix in (5).e. False. The statement is often true, but Span{u, v} is not a plane when v is a multiple of u, or whenu is the zero vector.24. a. True. See the beginning of the subsection Vectors in Rn.b. True. Use Fig. 7 to draw the parallelogram determined by u v and v.c. False. See the first paragraph of the subsection Linear Combinations.d. True. See the statement that refers to Fig. 11.e. True. See the paragraph following the definition of Span{v1, , vp}. 20. 20 CHAPTER 1 Linear Equations in Linear Algebra25. a. There are only three vectors in the set {a1, 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 methodof Exercise 13. 1 0 4 4 1 0 4 4 1 0 4 4 0 3 2 1 ~ 0 3 2 1 ~ 0 3 2 1 2 6 3 4 0 6 5 4 0 0 1 2 a 1 a 2 a 3bThe system for this augmented matrix is consistent, so b is in W.c. a1 = 1a1 + 0a2 + 0a3. See the discussion in the text following the definition of Span{v1, , vp}.26. a. [a1 a2 a3 b] =2 0 6 10 1 0 3 5 1 0 3 5 1 0 3 51 8 5 3 ~ 1 8 5 3 ~ 0 8 8 8 ~ 0 8 8 81 2 1 3 1 2 1 3 0 2 2 2 0 0 0 0 Yes, b is a linear combination of the columns of A, that is, b is in W.b. The third column of A is in W because a3 = 0a1 + 0a2 + 1a3.27. a. 5v1 is the output of 5 days operation of mine #1.v v .b. The total output is x1v1 + x2v2, so x1 and x2 should satisfy 1 1 2 21502825x x + = c. [M] Reduce the augmented matrix20 30 150 1 0 1.5 ~ 550 500 2825 0 1 4.0.Operate mine #1 for 1.5 days and mine #2 for 4 days. (This is the exact solution.)28. a. The amount of heat produced when the steam plant burns x1 tons of anthracite and x2 tons ofbituminous coal is 27.6x1 + 30.2x2 million Btu.b. The total output produced by x1 tons of anthracite and x2 tons of bituminous coal is given by the27.6 30.23100 6400250 360 + x xvector 1 2.27.6 30.2 1623100 6400 23,610250 360 1,623 + = c. [M] The appropriate values for x1 and x2 satisfy x 1 x2.To solve, row reduce the augmented matrix:27.6 30.2 162 1.000 0 3.9003100 6400 23610 ~ 0 1.000 1.800250 360 1623 0 0 0 The steam plant burned 3.9 tons of anthracite coal and 1.8 tons of bituminous coal. 21. 1.3 Solutions 2129. The total mass is 2 + 5 + 2 + 1 = 10. So v = (2v1 +5v2 + 2v3 + v4)/10. That is, + 5 4 4 9 10 20 8 9 1.31 2 4 5 3 2 3 8 1 8 15 6 8 .910 10= + + + = + + = 3 2 1 6 6 10 2 + 6 0v30. Let m be the total mass of the system. By definition,m m m mm m m1( ) 1kv = v +"+ v = v +"+ vk k k1 1 1The second expression displays v as a linear combination of v1, , vk, which shows that v is inSpan{v1, , vk}.31. a. The center of mass is + + = 1 0 8 2 10/ 3 1 1 13 1 1 4 2.b. The total mass of the new system is 9 grams. The three masses added, w1, w2, and w3, satisfy theequation + + + + + = 1 ( 0 w 1 1 ) ( 8 2 2 w 2 1 ) ( w3 1)9 1 1 4 2which can be rearranged to0 8 2 18 ( w ) ( ) ( ) 1 w 2 w3+ 1 + + 1 + + 1 = 1 1 4 18 and0 8 2 81 1 4 12 + + = w w w1 2 3The condition w1 + w2 + w3 = 6 and the vector equation above combine to produce a system of threeequations whose augmented matrix is shown below, along with a sequence of row operations:1 1 1 6 1 1 1 6 1 1 1 60 8 2 8 ~ 0 8 2 8 ~ 0 8 2 81 1 4 12 0 0 3 6 0 0 1 2 1 1 0 4 1 0 0 3.5 1 0 0 3.5~ 0 8 0 4 ~ 0 8 0 4 ~ 0 1 0 .50 0 1 2 0 0 1 2 0 0 1 2 Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4).Extra problem: Ignore the mass of the plate, and distribute 6 gm at the three vertices to make the center ofmass at (2, 2). Answer: Place 3 g at (0, 1), 1 g at (8, 1), and 2 g at (2, 4).32. See the parallelograms drawn on Fig. 15 from the text. Here c1, c2, c3, and c4 are suitable scalars. Thedarker parallelogram shows that b is a linear combination of v1 and v2, that isc1v1 + c2v2 + 0v3 = b 22. 22 CHAPTER 1 Linear Equations in Linear AlgebraThe larger parallelogram shows that b is a linear combination of v1 and v3, that is,v1 + 0v2 + c3v3 = bc4So the equation x1v1 + x2v2 + x3v3 = b has at least two solutions, not just one solution. (In fact, theequation has infinitely many solutions.)c2v2c3v30v3c4v1c1v1v1v2b33. a. For j = 1,, n, the jth entry of (u + v) + w is (uj + vj) + wj. By associativity of addition in R, thisentry equals uj + (vj + wj), which is the jth entry of u + (v + w). By definition of equality of vectors,(u + v) + w = u + (v + w).b. For any scalar c, the jth entry of c(u + v) is c(uj + vj), and the jth entry of cu + cv is cuj + cvj (bydefinition of scalar multiplication and vector addition). These entries are equal, by a distributive lawin R. So c(u + v) = cu + cv.34. a. For j = 1,, n, uj + (1)uj = (1)uj + uj = 0, by properties of R. By vector equality,u + (1)u = (1)u + u = 0.b. For scalars c and d, the jth entries of c(du) and (cd )u are c(duj) and (cd )uj, respectively. Theseentries in R are equal, so the vectors c(du) and (cd)u are equal.Note: When an exercise in this section involves a vector equation, the corresponding technology data (in thedata files on the web) is usually presented as a set of (column) vectors. To use MATLAB or other technology,a student must first construct an augmented matrix from these vectors. The MATLAB note in the Study Guidedescribes how to do this. The appendices in the Study Guide give corresponding information about Maple,Mathematica, and the TI and HP calculators.1.4 SOLUTIONSNotes: Key exercises are 120, 27, 28, 31 and 32. Exercises 29, 30, 33, and 34 are harder. Exercise 34anticipates the Invertible Matrix Theorem but is not used in the proof of that theorem.1. The matrix-vector product Ax product is not defined because the number of columns (2) in the 32matrix 4 2 1 6 0 1does not match the number of entries (3) in the vector327 . 23. 1.4 Solutions 232. The matrix-vector product Ax product is not defined because the number of columns (1) in the 31matrix261 does not match the number of entries (2) in the vector51 .3.6 5 6 5 12 15 324 3 2 4 3 3 8 9 137 6 7 6 14 18 4A = = = + = x , and6 5 62 5( 3) 324 3 ( 4) 2 ( 3) ( 3) 137 6 72 6( 3) 4A + = = + = + x4.18 3 4 8 3 4 8 3 4 71 1 1 15 1 2 5 1 2 5 1 2 81A + = = + + = = + + x , and18 3 4 81 31 ( 4) 1 715 1 2 51 11 21 81A + + = = = + + x5. On the left side of the matrix equation, use the entries in the vector x as the weights in a linearcombination of the columns of the matrix A: 5 1 8 4 8 + = 5 1 3 22 7 3 5 166. On the left side of the matrix equation, use the entries in the vector x as the weights in a linearcombination of the columns of the matrix A: 7 3 1 2 1 9 = 2 5 9 6 12 3 2 47. The left side of the equation is a linear combination of three vectors. Write the matrix A whose columnsare those three vectors, and create a variable vector x with three entries:4 5 7 4 5 71 3 8 1 3 87 5 0 7 5 04 1 2 4 1 2A = = andxxx 1= 2 3x . Thus the equation Ax = b is 4 5 7 6 x 1 3 8 1 8 x= 7 5 0 2 0 x 4 1 2 3 7 24. 24 CHAPTER 1 Linear Equations in Linear AlgebraFor your information: The unique solution of this equation is (5, 7, 3). Finding the solution by handwould be time-consuming.Note: The skill of writing a vector equation as a matrix equation will be important for both theory andapplication throughout the text. See also Exercises 27 and 28.8. The left side of the equation is a linear combination of four vectors. Write the matrix A whose columnsare those four vectors, and create a variable vector with four entries:4 4 5 3 4 4 5 32 5 4 0 2 5 4 0A = = , andzzzz 1 = 2 3 4z . Then the equation Az = biszzzz 1234 4 4 5 3 4 = 2 5 4 0 13 .For your information: One solution is (7, 3, 3, 1). The general solution is z1 = 6 + .75z3 1.25z4,z2 = 5 .5z3 .5z4, with z3 and z4 free.9. The system has the same solution set as the vector equationx x x 3 1 5 90 1 4 0 + + = 1 2 3and this equation has the same solution set as the matrix equationxxx 123 3 1 5 9 0 1 4 = 0 10. The system has the same solution set as the vector equation 8 1 4 5 + 4 = 1 1 3 2x x1 2and this equation has the same solution set as the matrix equation 8 1 4 x1 5 4 = 1 1 3 x2 2 11. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system: 1 2 4 2 1 2 4 2 1 2 4 2 1 2 0 6 1 0 0 0 0 1 5 2 ~ 0 1 5 2 ~ 0 1 5 2 ~ 0 1 0 3 ~ 0 1 0 3 2 4 3 9 0 0 5 5 0 0 1 1 0 0 1 1 0 0 1 1 25. 1.4 Solutions 25The solution is123031xxx= = =. As a vector, the solution is x =123031xxx = .12. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system:1 2 1 0 1 2 1 0 1 2 1 0 1 2 1 03 1 2 1 ~ 0 5 5 1 ~ 0 5 5 1 ~ 0 5 5 10 5 3 1 0 5 3 1 0 0 2 2 0 0 1 1 1 2 0 1 1 2 0 1 1 0 0 3/5~ 0 5 0 4 ~ 0 1 0 4/5 ~ 0 1 0 4/50 0 1 1 0 0 1 1 0 0 1 1 The solution is1233/54/51xxx= = =. As a vector, the solution is x =1233/54/51xxx = .13. The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of thecolumns of A. This happens if and only if the equation Ax = u has a solution. (See the box precedingExample 3 in Section 1.4.) To study this equation, reduce the augmented matrix [A u]3 5 0 1 1 4 1 1 4 1 1 42 6 4 ~ 2 6 4 ~ 0 8 12 ~ 0 8 121 1 4 3 5 0 0 8 12 0 0 0 The equation Ax = u has a solution, so u is in the plane spanned by the columns of A.For your information: The unique solution of Ax = u is (5/2, 3/2).14. Reduce the augmented matrix [A u] to echelon form:5 8 7 2 1 3 0 2 1 3 0 2 1 3 0 20 1 1 3 ~ 0 1 1 3 ~ 0 1 1 3 ~ 0 1 1 31 3 0 2 5 8 7 2 0 7 7 8 0 0 0 29 The equation Ax = u has no solution, so u is not in the subset spanned by the columns of A.15. The augmented matrix for Ax = b is 122 16 3bb bb b 2 1 0 0 + 3, which is row equivalent to 12 1.This shows that the equation Ax = b is not consistent when 3b1 + b2 is nonzero. The set of b for which theequation is consistent is a line through the originthe set of all points (b1, b2) satisfying b2 = 3b1.16. Row reduce the augmented matrix [A b]:b 1 3 43 2 6, .5 1 8123 = = A bbbb bb b bb b b 1 3 4 1 1 3 4 1 3 2 6 2 ~ 0 7 6 2 + 31 5 1 8 3 0 14 12 3 51 26. 26 CHAPTER 1 Linear Equations in Linear Algebrab bb b b b 1 3 4 1 1 3 4~ 1 0 7 6 2 + 3 1 = 0 7 6 2 + 3 1 0 0 0 b 3 5 b + 2( b + 3 b 1 2 1 ) 0 0 0 b 1 + 2b 2 + b3The equation Ax = b is consistent if and only if b1 + 2b2 + b3 = 0. The set of such b is a plane through theorigin in R3.17. Row reduction shows that only three rows of A contain a pivot position:1 3 0 3 1 3 0 3 1 3 0 3 1 3 0 31 1 1 1 0 2 1 4 0 2 1 4 0 2 1 4 A= ~ ~ ~ 0 4 2 8 0 4 2 8 0 0 0 0 0 0 0 5 2 0 3 1 0 6 3 7 0 0 0 5 0 0 0 0Because not every row of A contains a pivot position, Theorem 4 in Section 1.4 shows that the equationAx = b does not have a solution for each b in R4.18. Row reduction shows that only three rows of B contain a pivot position: 1 3 2 2 1 3 2 2 1 3 2 2 1 3 2 2 0 1 1 5 0 1 1 5 0 1 1 5 0 1 1 5 B= ~ ~ ~ 1 2 3 7 0 1 1 5 0 0 0 0 0 0 0 7 2 8 2 1 0 2 2 3 0 0 0 7 0 0 0 0Because not every row of B contains a pivot position, Theorem 4 in Section 1.4 shows that the equationBx = y does not have a solution for each y in R4.19. The work in Exercise 17 shows that statement (d) in Theorem 4 is false. So all four statements inTheorem 4 are false. Thus, not all vectors in R4 can be written as a linear combination of the columnsof A. Also, the columns of A do not span R4.20. The work in Exercise 18 shows that statement (d) in Theorem 4 is false. So all four statements inTheorem 4 are false. Thus, not all vectors in R4 can be written as a linear combination of the columnsof B. The columns of B certainly do not span R3, because each column of B is in R4, not R3. (Thisquestion was asked to alert students to a fairly common misconception among students who are justlearning about spanning.)21. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row.1 0 1 1 0 1 1 0 1 1 0 10 1 0 0 1 0 0 1 0 0 1 0 ~ ~ ~ . 1 0 0 0 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 0 0 The matrix [v1 v2 v3] does not have a pivot in each row, so the columns of the matrix do not span R4,by Theorem 4. That is, {v1, v2, v3} does not span R4.Note: Some students may realize that row operations are not needed, and thereby discover the principlecovered in Exercises 31 and 32. 27. 1.4 Solutions 2722. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row. 0 0 4 2 8 5 0 3 1 ~ 0 3 1 2 8 5 0 0 4The matrix [v1 v2 v3] has a pivot in each row, so the columns of the matrix span R4, by Theorem 4.That is, {v1, v2, v3} spans R4.23. a. False. See the paragraph following equation (3). The text calls Ax = b a matrix equation.b. True. See the box before Example 3.c. False. See the warning following Theorem 4.d. True. See Example 4.e. True. See parts (c) and (a) in Theorem 4.f. True. In Theorem 4, statement (a) is false if and only if statement (d) is also false.24. a. True. This statement is in Theorem 3. However, the statement is true without any "proof" because, bydefinition, Ax is simply a notation for x1a1 + + xnan, where a1, , an are the columns of A.b. True. See Example 2.c. True, by Theorem 3.d. True. See the box before Example 2. Saying that b is not in the set spanned by the columns of A is thesame a saying that b is not a linear combination of the columns of A.e. False. See the warning that follows Theorem 4.f. True. In Theorem 4, statement (c) is false if and only if statement (a) is also false.25. By definition, the matrix-vector product on the left is a linear combination of the columns of the matrix,in this case using weights 3, 1, and 2. So c1 = 3, c2 = 1, and c3 = 2.26. The equation in x1 and x2 involves the vectors u, v, and w, and it may be viewed as[ ] 12.xx = u v w By definition of a matrix-vector product, x1u + x2v = w. The stated fact that3u 5v w = 0 can be rewritten as 3u 5v = w. So, a solution is x1 = 3, x2 = 5.27. Place the vectors q1, q2, and q3 into the columns of a matrix, say, Q and place the weights x1, x2, and x3into a vector, say, x. Then the vector equation becomesQx = v, where Q = [q1 q2 q3] andxxx 1= 2 3xNote: If your answer is the equation Ax = b, you need to specify what A and b are.28. The matrix equation can be written as c1v1 + c2v2 + c3v3 + c4v4 + c5v5 = v6, wherec1= 3, c2 = 2, c3 = 4, c4 = 1, c5 = 2, and 3 5 4 9 7 8= , = , = , = , = ,= 5 8 1 2 4 1v v v v v v1 2 3 4 5 6 28. 28 CHAPTER 1 Linear Equations in Linear Algebra29. Start with any 33 matrix B in echelon form that has three pivot positions. Perform a row operation(a row interchange or a row replacement) that creates a matrix A that is not in echelon form. Then A hasthe desired property. The justification is given by row reducing A to B, in order to display the pivotpositions. Since A has a pivot position in every row, the columns of A span R3, by Theorem 4.30. Start with any nonzero 33 matrix B in echelon form that has fewer than three pivot positions. Performa row operation that creates a matrix A that is not in echelon form. Then A has the desired property. SinceA does not have a pivot position in every row, the columns of A do not span R3, by Theorem 4.31. A 32 matrix has three rows and two columns. With only two columns, A can have at most two pivotcolumns, and so A has at most two pivot positions, which is not enough to fill all three rows. ByTheorem 4, the equation Ax = b cannot be consistent for all b in R3. Generally, if A is an mn matrixwith m > n, then A can have at most n pivot positions, which is not enough to fill all m rows. Thus, theequation Ax = b cannot be consistent for all b in R3.32. A set of three vectors in cannot span R4. Reason: the matrix A whose columns are these three vectors hasfour rows. To have a pivot in each row, A would have to have at least four columns (one for each pivot),which is not the case. Since A does not have a pivot in every row, its columns do not span R4, byTheorem 4. In general, a set of n vectors in Rm cannot span Rm when n is less than m.33. If the equation Ax = b has a unique solution, then the associated system of equations does not have anyfree variables. If every variable is a basic variable, then each column of A is a pivot column. So thereduced echelon form of A must be 1 0 00 1 0 0 0 10 0 0.Note: Exercises 33 and 34 are difficult in the context of this section because the focus in Section 1.4 is onexistence of solutions, not uniqueness. However, these exercises serve to review ideas from Section 1.2, andthey anticipate ideas that will come later.34. If the equation Ax = b has a unique solution, then the associated system of equations does not have anyfree variables. If every variable is a basic variable, then each column of A is a pivot column. So thereduced echelon form of A must be 1 0 00 1 0 0 0 1. Now it is clear that A has a pivot position in each row.By Theorem 4, the columns of A span R3.35. Given Ax1 = y1 and Ax2 = y2, you are asked to show that the equation Ax = w has a solution, wherew = y1 + y2. Observe that w = Ax1 + Ax2 and use Theorem 5(a) with x1 and x2 in place of u and v,respectively. That is, w = Ax1 + Ax2 = A(x1 + x2). So the vector x = x1 + x2 is a solution of w = Ax.36. Suppose that y and z satisfy Ay = z. Then 4z = 4Ay. By Theorem 5(b), 4Ay = A(4y). So 4z = A(4y),which shows that 4y is a solution of Ax = 4z. Thus, the equation Ax = 4z is consistent.37. [M] 7 25 8 7 25 8 7 25 8 53 49011/7 3/723/7 011/7 3/723/7 ~ ~ 6 102 7 0 58/ 7 16/ 7 1/ 7 0 0 50/11189/11 7 9 2 15 0 113 23 0 0 0 0 29. 1.4 Solutions 29or, approximately 7 2 5 8 0 1.57 .429 3.29 0 0 4.55 17.2 0 0 0 0, to three significant figures. The original matrix does nothave a pivot in every row, so its columns do not span R4, by Theorem 4.38. [M] 5 7 4 9 5 7 4 9 5 7 4 9 6 8 7 5 0 2/ 5 11/5 29/ 5 0 2/5 11/ 5 29/5 ~ ~ 4 4 9 9 0 8/5 29/5 81/5 0 0 3 7 9 11 16 7 0 8/ 5 44/ 5 116/ 5 0 0 * * MATLAB shows starred entries for numbers that are essentially zero (to many decimal places). So, withpivots only in the first three rows, the original matrix has columns that do not span R4, by Theorem 4. 12 7 11 9 5 12 7 11 9 50 5/ 4 1/4 1/4 3/4 0 5/4 1/ 4 1/4 3/ 4~ ~39. [M]12 7 11 9 5 12 7 11 9 59 4 8 7 3 0 5/4 1/ 4 1/4 3/ 4~6 11 7 3 9 0 15/2 3/2 3/2 13/24 6 10 5 12 0 11/ 3 19/ 3 2 31/ 3 0 0 0 0 2 0 0 28/ 5 41/15 122/15 0 0 28/ 5 41/15 122/15 0 0 0 0 2 The original matrix has a pivot in every row, so its columns span R4, by Theorem 4. 8 11 6 7 13 8 11 6 7 130 13/8 1/ 4 1/8 19/8 0 13/8 1/ 4 1/8 19/8~ ~0 0 0 0 12 0 0 0 6 00 0 0 6 0 0 0 0 0 1240. [M]8 11 6 7 13 8 11 6 7 137 8 5 6 9 0 13/8 1/ 4 1/8 19/8~11 7 7 9 6 0 65/8 5/ 4 5/8 191/83 4 1 8 7 0 65/8 5/4 43/8 95/8 The original matrix has a pivot in every row, so its columns span R4, by Theorem 4.41. [M] Examine the calculations in Exercise 39. Notice that the fourth column of the original matrix, say A,is not a pivot column. Let Ao be the matrix formed by deleting column 4 of A, let B be the echelon formobtained from A, and let Bo be the matrix obtained by deleting column 4 of B. The sequence of rowoperations that reduces A to B also reduces Ao to Bo. Since Bo is in echelon form, it shows that Ao has apivot position in each row. Therefore, the columns of Ao span R4.It is possible to delete column 3 of A instead of column 4. In this case, the fourth column of A becomes apivot column of Ao, as you can see by looking at what happens when column 3 of B is deleted. For laterwork, it is desirable to delete a nonpivot column. 30. 30 CHAPTER 1 Linear Equations in Linear AlgebraNote: Exercises 41 and 42 help to prepare for later work on the column space of a matrix. (See Section 2.9 or4.6.) The Study Guide points out that these exercises depend on the following idea, not explicitly mentionedin the text: when a row operation is performed on a matrix A, the calculations for each new entry depend onlyon the other entries in the same column. If a column of A is removed, forming a new matrix, the absence ofthis column has no affect on any row-operation calculations for entries in the other columns of A. (Theabsence of a column might affect the particular choice of row operations performed for some purpose, but thatis not being considered here.)42. [M] Examine the calculations in Exercise 40. The third column of the original matrix, say A, is not apivot column. Let Ao be the matrix formed by deleting column 3 of A, let B be the echelon form obtainedfrom A, and let Bo be the matrix obtained by deleting column 3 of B. The sequence of row operations thatreduces A to B also reduces Ao to Bo. Since Bo is in echelon form, it shows that Ao has a pivot position ineach row. Therefore, the columns of Ao span R4.It is possible to delete column 2 of A instead of column 3. (See the remark for Exercise 41.) However,only one column can be deleted. If two or more columns were deleted from A, the resulting matrix wouldhave fewer than four columns, so it would have fewer than four pivot positions. In such a case, not everyrow could contain a pivot position, and the columns of the matrix would not span R4, by Theorem 4.Notes: At the end of Section 1.4, the Study Guide gives students a method for learning and mastering linearalgebra concepts. Specific directions are given for constructing a review sheet that connects the basicdefinition of span with related ideas: equivalent descriptions, theorems, geometric interpretations, specialcases, algorithms, and typical computations. I require my students to prepare such a sheet that reflects theirchoices of material connected with span, and I make comments on their sheets to help them refine theirreview. Later, the students use these sheets when studying for exams.The MATLAB box for Section 1.4 introduces two useful commands gauss and bgauss that allow astudent to speed up row reduction while still visualizing all the steps involved. The commandB = gauss(A,1) causes MATLAB to find the left-most nonzero entry in row 1 of matrix A, and use thatentry as a pivot to create zeros in the entries below, using row replacement operations. The result is a matrixthat a student might write next to A as the first stage of row reduction, since there is no need to write a newmatrix after each separate row replacement. I use the gauss command frequently in lectures to obtain anechelon form that provides data for solving various problems. For instance, if a matrix has 5 rows, and if rowswaps are not needed, the following commands produce an echelon form of A:B = gauss(A,1), B = gauss(B,2), B = gauss(B,3), B = gauss(B,4)If an interchange is required, I can insert a command such as B = swap(B,2,5) . The command bgaussuses the left-most nonzero entry in a row to produce zeros above that entry. This command, together withscale, can change an echelon form into reduced echelon form.The use of gauss and bgauss creates an environment in which students use their computer programthe same way they work a problem by hand on an exam. Unless you are able to conduct your exams in acomputer laboratory, it may be unwise to give students too early the power to obtain reduced echelon formswith one commandthey may have difficulty performing row reduction by hand during an exam. Instructorswhose students use a graphic calculator in class each day do not face this problem. In such a case, you maywish to introduce rref earlier in the course than Chapter 4 (or Section 2.8), which is where I finally allowstudents to use that command.1.5 SOLUTIONSNotes: The geometry helps students understand Span{u, v}, in preparation for later discussions of subspaces.The parametric vector form of a solution set will be used throughout the text. Figure 6 will appear again inSections 2.9 and 4.8. 31. 1.5 Solutions 31For solving homogeneous systems, the text recommends working with the augmented matrix, although nocalculations take place in the augmented column. See the Study Guide comments on Exercise 7 that illustratetwo common student errors.All students need the practice of Exercises 114. (Assign all odd, all even, or a mixture. If you do notassign Exercise 7, be sure to assign both 8 and 10.) Otherwise, a few students may be unable later to find abasis for a null space or an eigenspace. Exercises 2934 are important. Exercises 33 and 34 help students laterunderstand how solutions of Ax = 0 encode linear dependence relations among the columns of A. Exercises3538 are more challenging. Exercise 37 will help students avoid the standard mistake of forgetting thatTheorem 6 applies only to a consistent equation Ax = b.1. Reduce the augmented matrix to echelon form and circle the pivot positions. If a column of thecoefficient matrix is not a pivot column, the corresponding variable is free and the system of equationshas a nontrivial solution. Otherwise, the system has only the trivial solution.2 5 8 0 2 5 8 0 2 5 8 02 7 1 0 ~ 0 12 9 0 ~ 0 12 9 04 2 7 0 0 12 9 0 0 0 0 0 The variable x3 is free, so the system has a nontrivial solution.2. 1 3 7 0 1 3 7 0 1 3 7 0 2 1 4 0 ~ 0 5 10 0 ~ 0 5 10 0 1 2 9 0 0 5 2 0 0 0 12 0There is no free variable; the system has only the trivial solution.3. 3 5 7 0 3 5 7 0~ 6 7 1 0 0 3 15 0 . The variable x3 is free; the system has nontrivial solutions.An alert student will realize that row operations are unnecessary. With only two equations, there can beat most two basic variables. One variable must be free. Refer to Exercise 31 in Section 1.2.4. 5 7 9 0 1 2 6 0 1 2 6 0 ~ ~ 1 2 6 0 5 7 9 0 0 3 39 0 . x3 is a free variable; the system hasnontrivial solutions. As in Exercise 3, row operations are unnecessary.5. 1 3 1 0 1 3 1 0 1 0 5 0 1 0 5 0 4 9 2 0 ~ 0 3 6 0 ~ 0 3 6 0 ~ 0 1 2 0 0 3 6 0 0 3 6 0 0 0 0 0 0 0 0 0x xx x =+ =5 02 00 01 32 3=. The variable x3 is free, x1 = 5x3, and x2 = 2x3.In parametric vector form, the general solution is5 52 21 32 3 33 31x xx x xx x = = = x . 32. 32 CHAPTER 1 Linear Equations in Linear Algebra6. 1 3 5 0 1 3 5 0 1 3 5 0 1 0 4 0 1 4 8 0 ~ 0 1 3 0 ~ 0 1 3 0 ~ 0 1 3 0 3 7 9 0 0 2 6 0 0 0 0 0 0 0 0 0x xx x+ = =4 03 00 01 32 3=. The variable x3 is free, x1 = 4x3, and x2 = 3x3.In parametric vector form, the general solution is4 43 31 32 3 33 31x xx x xx x = = = x .7. 1 3 3 7 0 1 0 9 8 0 ~ 0 1 4 5 0 0 1 4 5 0 x x xx x x+ = + =9 8 04 5 0. 1 3 42 3 4The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = 9x3 + 8x4, and x2 = 4x3 5x4. Thegeneral solution isx x x x xx x x x x 1 9 3 + 8 4 9 3 8 4 9 84 5 4 2 3 4 3 5 4 5 = = = + 4 = x + x x x x 0 3 1 4 3 3 30 x x 0 x 0 4 4 4 1x8. 1 2 9 5 0 1 0 5 7 0 0 1 2 6 0 ~ 0 1 2 6 0 x x xx x x =+ =5 7 02 6 0. 1 3 42 3 4The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = 5x3 + 7x4 and x2 = 2x3 + 6x4. The generalsolution in parametric vector form isx x x x xx x x x x 1 5 3 + 7 4 5 3 7 4 5 7 2 + 6 2 6 2 = 2 = 3 4 = 3 + 4 = x 63 + x x x x 0 1 4 3 3 30x x 0 x 0 4 4 4 1x9. 3 9 6 0 1 3 2 0 1 3 2 0~ ~ 1 3 2 0 3 9 6 0 0 0 0 0 x x + x =1 2 3 3 2 0=0 0.The solution is x1 = 3x2 2x3, with x2 and x3 free. In parametric vector form,x x x xx x x xx x 3 2 2 3 3 2 2 3 3 2= = + = + x 0 1 0.2 2 2 33 30 0 110. 1 3 0 4 0 1 3 0 4 0 ~ 2 6 0 8 0 0 0 0 0 0 x x x =1 2 4 3 4 0=0 0.The only basic variable is x1, so x2, x3, and x4 are free. (Note that x3 is not zero.) Also, x1 = 3x2 + 4x4. Thegeneral solution is 33. 1.5 Solutions 33x x x x xx x x 1 3 2 + 4 4 3 2 0 4 3 0 4 4 2 2 2 0 0 1 0 0= = = + + = x 2 + x + x x x 0 x 0 0 3 1 4 3 0 3 3 x 4 x 4 0 0 x 4 0 0 1x11. 1 4 2 0 3 5 0 1 4 2 0 0 7 0 1 4 0 0 0 5 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 ~ ~ 0 0 0 0 1 4 0 0 0 0 0 1 4 0 0 0 0 0 1 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0x x x + =4 5 01 2 6x x = =3 6x x5 604 00 =0. The basic variables are x1, x3, and x5. The remaining variables are free.In particular, x4 is free (and not zero as some may assume). The solution is x1 = 4x2 5x6, x3 = x6,x5 = 4x6, with x2, x4, and x6 free. In parametric vector form,x x x x xx x xx x x = = = + + = + 4 5 4 0 5 4 01 2 6 2 62 2 23 6 6 + 0 0 1 00 0 0 00 0 0x x x62 4x x xx x xx x x4 4 45 6 66 6 64 0 0 4 00 0 0501x1 00 40 1 u v wNote: The Study Guide discusses two mistakes that students often make on this type of problem.12. 1 5 2 6 9 0 0 1 5 2 6 9 0 0 1 5 0 8 1 0 0 0 0 1 7 4 8 0 0 0 1 7 4 0 0 0 0 1 7 4 0 0 ~ ~ 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0+ + + =5 8 01 2 4 5 + =7 4 03 4 5600 0x x x xx x xx==.The basic variables are x1, x3, and x6; the free variables are x2, x4, and x5. The general solution isx1 = 5x2 8x4 x5, x3 = 7x4 4x5, and x6 = 0. In parametric vector form, the solution is 34. 34 CHAPTER 1 Linear Equations in Linear Algebrax x x x x x xx x xx x x x x 1 5 2 8 4 5 5 2 8 4 5 5 2 2 2 0 0 1 3 7 4 4 5 0 7 4 4 50= = = + + = + + x x24 5x x xx x xx0 0 00 0 04 4 45 5 560 0 0 0 08 10 07 41 00 10 0x x13. To write the general solution in parametric vector form, pull out the constant terms that do not involvethe free variable:x x xx x x x xx x x + = = = + = + = + 5 4 5 4 5 42 7 2 7 2 7 .1 3 32 3 3 3 33 3 3x pq0 0 1 p qGeometrically, the solution set is the line through520 in the direction of471 .14. To write the general solution in parametric vector form, pull out the constant terms that do not involvethe free variable:x x xx x x3 0 3 0 38 8 8 12 5 2 5 2 5 1 4 4 + = 2 = 4 = + 4 = + = + x pqx x4 4x x xx x x3 4 44 4 40 0 1 p qThe solution set is the line through p in the direction of q.15. Row reduce the augmented matrix for the system:1 3 1 1 1 3 1 1 1 3 1 14 9 2 1 ~ 0 3 6 3 ~ 0 3 6 30 3 6 3 0 3 6 3 0 0 0 0 1 3 1 1 1 0 5 2~ 0 1 2 1 ~ 0 1 2 10 0 0 0 0 0 0 0 .x xx x =+ =5 22 10 01 32 3=.Thus x1 = 2 + 5x3, x2 = 1 2x3, and x3 is free. In parametric vector form,x x xx x x xx x x 1 2 + 5 3 2 5 3 2 5= = 2 1 2 3 = 1 + 2 3 = 1 + 3 2 0 0 3 3 3 1x 35. 1.5 Solutions 35The solution set is the line through 2 1 0, parallel to the line that is the solution set of the homogeneoussystem in Exercise 5.16. Row reduce the augmented matrix for the system: 1 3 5 4 1 3 5 4 1 3 5 4 1 0 4 5 1 4 8 7 ~ 0 1 3 3 ~ 0 1 3 3 ~ 0 1 3 3 3 7 9 6 0 2 6 6 0 0 0 0 0 0 0 0x xx x+ = =4 53 30 01 32 3=. Thus x1 = 5 4x3, x2 = 3 + 3x3, and x3 is free. In parametric vector form,x x xx x x xx x x 1 5 4 3 5 4 3 5 4= 2 = 3 + 3 = 3 + 3 = 3 3 3 + 3 3 0 0 3 3 3 1xThe solution set is the line through 5 3 0, parallel to the line that is the solution set of the homogeneoussystem in Exercise 6.17. Solve x1 + 9x2 4x3 = 2 for the basic variable: x1 = 2 9x2 + 4x3, with x2 and x3 free. In vector form,the solution isx x x x xx x x x xx x x 1 2 9 2 + 4 3 2 9 2 4 3 2 9 4= = = 0 + + 0 = 0 + 2 2 2 2 1 + 3 0 0 3 3 0 3 0 0 1xThe solution of x1 + 9x2 4x3 = 0 is x1 = 9x2 + 4x3, with x2 and x3 free. In vector form,x x x x xx x x x xx x x 1 9 2 + 4 3 9 2 4 3 9 4= 2 = 2 = 2 + = 2 + 3 3 3 3 x 0 1 0= x2u + x3v0 0 1The solution set of the homogeneous equation is the plane through the origin in R3 spanned byu and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through thepoint p = 2 0 0.18. Solve x1 3x2 + 5x3 = 4 for the basic variable: x1 = 4 + 3x2 5x3, with x2 and x3 free. In vector form, thesolution isx x x x xx x x x xx x x 1 4 + 3 2 5 3 4 3 2 5 3 4 3 5= = = 0 2 2 + 2 + 0 = 0 + 2 1 + 3 0 3 0 0 3 3 0 0 1x 36. 36 CHAPTER 1 Linear Equations in Linear AlgebraThe solution of x1 3x2 + 5x3 = 0 is x1 = 3x2 5x3, with x2 and x3 free. In vector form,x x x x xx x x x xx x x 1 3 2 5 3 3 2 5 3 3 5= = 2 2 = 2 + = 2 + 3 3 3 3 x 0 1 0= x2u + x3v0 0 1The solution set of the homogeneous equation is the plane through the origin in R3 spanned by u and v.The solution set of the nonhomogeneous equation is parallel to this plane and passes through thepoint p =400 .19. The line through a parallel to b can be written as x = a + t b, where t represents a parameter: = + x = 122 50 3xtxx tx t= =, or 122 5320. The line through a parallel to b can be written as x = a + tb, where t represents a parameter: = + x = 123 74 8xtxx tx t= = +, or 123 74 821. The line through p and q is parallel to q p. So, given 2 3= and= 5 1 p q , form 3 2 51 ( 5) 6q p , and write the line as x = p + t(q p) = = = 2 5 + t 5 6 .22. The line through p and q is parallel to q p. So, given 6 0= = p q , formand3 4 0 ( 6) 64 3 7q p , and write the line as x = p + t(q p) = = = 6 6 + t 3 7Note: Exercises 21 and 22 prepare for Exercise 27 in Section 1.8.23. a. True. See the first paragraph of the subsection titled Homogeneous Linear Systems.b. False. The equation Ax = 0 gives an implicit description of its solution set. See the subsection entitledParametric Vector Form.c. False. The equation Ax = 0 always has the trivial solution. The box before Example 1 uses the wordnontrivial instead of trivial.d. False. The line goes through p parallel to v. See the paragraph that precedes Fig. 5.e. False. The solution set could be empty! The statement (from Theorem 6) is true only when thereexists a vector p such that Ap = b.24. a. False. A nontrivial solution of Ax = 0 is any nonzero x that satisfies the equation. See thesentence before Example 2.b. True. See Example 2 and the paragraph following it. 37. 1.5 Solutions 37c. True. If the zero vector is a solution, then b = Ax = A0 = 0.d. True. See the paragraph following Example 3.e. False. The statement is true only when the solution set of Ax = 0 is nonempty. Theorem 6 appliesonly to a consistent system.25. Suppose p satisfies Ax = b. Then Ap = b. Theorem 6 says that the solution set of Ax = b equals theset S ={w : w = p + vh for some vh such that Avh = 0}. There are two things to prove: (a) every vectorin S satisfies Ax = b, (b) every vector that satisfies Ax = b is in S.a. Let w have the form w = p + vh, where Avh = 0. ThenAw = A(p + vh) = Ap + Avh. By Theorem 5(a) in section 1.4= b + 0 = bSo every vector of the form p + vh satisfies Ax = b.b. Now let w be any solution of Ax = b, and set vh = w p. ThenAvh = A(w p) = Aw Ap = b b = 0So vh satisfies Ax = 0. Thus every solution of Ax = b has the form w = p + vh.26. (Geometric argument using Theorem 6.) Since Ax = b is consistent, its solution set is obtained bytranslating the solution set of Ax = 0, by Theorem 6. So the solution set of Ax = b is a single vector ifand only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has onlythe trivial solution.(Proof using free variables.) If Ax = b has a solution, then the solution is unique if and only if thereare no free variables in the corresponding system of equations, that is, if and only if every column of A isa pivot column. This happens if and only if the equation Ax = 0 has only the trivial solution.27. When A is the 33 zero matrix, every x in R3 satisfies Ax = 0. So the solution set is all vectors in R3.28. No. If the solution set of Ax = b contained the origin, then 0 would satisfy A0= b, which is not truesince b is not the zero vector.29. a. When A is a 33 matrix with three pivot positions, the equation Ax = 0 has no free variables andhence has no nontrivial solution.b. With three pivot positions, A has a pivot position in each of its three rows. By Theorem 4 inSection 1.4, the equation Ax = b has a solution for every possible b. The term possible in theexercise means that the only vectors considered in this case are those in R3, because A has three rows.30. a. When A is a 33 matrix with two pivot positions, the equation Ax = 0 has two basic variables andone free variable. So Ax = 0 has a nontrivial solution.b. With only two pivot positions, A cannot have a pivot in every row, so by Theorem 4 in Section 1.4,the equation Ax = b cannot have a solution for every possible b (in R3).31. a. When A is a 32 matrix with two pivot positions, each column is a pivot column. So the equationAx = 0 has no free variables and hence no nontrivial solution.b. With two pivot positions and three rows, A cannot have a pivot in every row. So the equation Ax = bcannot have a solution for every possible b (in R3), by Theorem 4 in Section 1.4.32. a. When A is a 24 matrix with two pivot positions, the equation Ax = 0 has two basic variables andtwo free variables. So Ax = 0 has a nontrivial solution.b. With two pivot positions and only two rows, A has a pivot position in every row. By Theorem 4 inSection 1.4, the equation Ax = b has a solution for every possible b (in R2). 38. 38 CHAPTER 1 Linear Equations in Linear Algebra 2 6 7 + 21 3 933. Look at x 1 x2and notice that the second column is 3 times the first. So suitable values forx1 and x2 would be 3 and 1 respectively. (Another pair would be 6 and 2, etc.) Thus31 = xsatisfies Ax = 0.34. Inspect how the columns a1 and a2 of A are related. The second column is 3/2 times the first. Putanother way, 3a1 + 2a2 = 0. Thus32 satisfies Ax = 0.Note: Exercises 33 and 34 set the stage for the concept of linear dependence.35. Look for A = [a1 a2 a3] such that 1a1 + 1a2 + 1a3 = 0. That is, construct A so that each row sum (thesum of the entries in a row) is zero.36. Look for A = [a1 a2 a3] such that 1a1 2a2 + 1a3 = 0. That is, construct A so that the sum of thefirst and third columns is twice the second column.37. Since the solution set of Ax = 0 contains the point (4,1), the vector x = (4,1) satisfies Ax = 0. Write thisequation as a vector equation, using a1 and a2 for the columns of A:4a1 + 1a2 = 0Then a2 = 4a1. So choose any nonzero vector for the first column of A and multiply that column by 4to get the second column of A. For example, set1 41 4A = .Finally, the only way the solution set of Ax = b could not be parallel to the line through (1,4) and theorigin is for the solution set of Ax = b to be empty. This does not contradict Theorem 6, because thattheorem applies only to the case when the equation Ax = b has a nonempty solution set. For b, take anyvector that is not a multiple of the columns of A.Note: In the Study Guide, a Checkpoint for Section 1.5 will help students with Exercise 37.38. No. If Ax = y has no solution, then A cannot have a pivot in each row. Since A is 33, it has at most twopivot positions. So the equation Ax = z for any z has at most two basic variables and at least one freevariable. Thus, the solution set for Ax = z is either empty or has infinitely many elements.39. If u satisfies Ax = 0, then Au = 0. For any scalar c, Theorem 5(b) in Section 1.4 shows that A(cu) =cAu = c0 = 0.40. Suppose Au = 0 and Av = 0. Then, since A(u + v) = Au + Av by Theorem 5(a) in Section 1.4,A(u + v) = Au + Av = 0 + 0 = 0.Now, let c and d be scalars. Using both parts of Theorem 5,A(cu + dv) = A(cu) + A(dv) = cAu + dAv = c0 + d0 = 0.Note: The MATLAB box in the Study Guide introduces the zeros command, in order to augment a matrixwith a column of zeros. 39. 1.6 Solutions 391.6 SOLUTIONS1. Fill in the exchange table one column at a time. The entries in a column describe where a sector's outputgoes. The decimal fractions in each column sum to 1.Distribution ofOutput From:Goods Services Purchased by: output input.2 .7 Goods.8 .3 ServicesDenote the total annual output (in dollars) of the sectors by pG and pS. From the first row, the total inputto the Goods sector is .2 pG + .7 pS. The Goods sector must pay for that. So the equilibrium prices mustsatisfyincome expensesp = .2p + .7 pG G SFrom the second row, the input (that is, the expense) of the Services sector is .8 pG + .3 pS.The equilibrium equation for the Services sector isincome expensesp = .8p + .3pS G SMove all variables to the left side and combine like terms:p pp p =.8 .7 0.8 .7 0G SG S + =Row reduce the augmented matrix:.8 .7 0 .8 .7 0 1 .875 0 ~ ~ .8 .7 0 0 0 0 0 0 0 The general solution is pG = .875 pS, with pS free. One equilibrium solution is pS = 1000 and pG = 875.If one uses fractions instead of decimals in the calculations, the general solution would be writtenpG = (7/8) pS, and a natural choice of prices might be pS = 80 and pG = 70. Only the ratio of the pricesis important: pG = .875 pS. The economic equilibrium is unaffected by a proportional change in prices.2. Take some other value for pS, say 200 million dollars. The other equilibrium prices are thenpC = 188 million, pE = 170 million. Any constant nonnegative multiple of these prices is a set ofequilibrium prices, because the solution set of the system of equations consists of all multiples of onevector. Changing the unit of measurement to, say, European euros has the same effect as multiplyingall equilibrium prices by a constant. The ratios of the prices remain the same, no matter what currencyis used.3. a. Fill in the exchange table one column at a time. The entries in a column describe where a sectorsoutput goes. The decimal fractions in each column sum to 1. 40. 40 CHAPTER 1 Linear Equations in Linear AlgebraDistribution of Output From: PurchasedChemicals Fuels Machinery by:output input.2 .8 .4 Chemicals.3 .1 .4 Fuels.5 .1 .2 Machineryb. Denote the total annual output (in dollars) of the sectors by pC, pF, and pM. From the first row of thetable, the total input to the ChemicalMetals sector is .2 pC + .8 pF + .4 pM. So the equillibriumprices must satisfyincome expensesp = .2p + .8p + .4 pC C F MFrom the second and third rows of the table, the income/expense requirements for the FuelsPowersector and the Machinery sector are, respectively,p = .3 p + .1 p +.4pp = .5 p + .1 p +.2pF C F MM C F MMove all variables to the left side and combine like terms:p p pp p pp p p.8 .8 .4 0.3 .9 .4 0.5 .1 .8 0C F MC F MC F M=+ =+ =c. [M] You can obtain the reduced echelon form with a matrix program. Actually, hand calculations arenot too messy. To simplify the calculations, first scale each row of the augmented matrix by 10, thencontinue as usual. rary.8 8 4 0 1 1 .5 0 1 1 .5 03 9 4 0 ~ 3 9 4 0 ~ 0 6 5.5 05 1 8 0 5 1 8 0 0 6 5.5 01 1 .5 0 1 0 1.417 0 The number of decimal~ 0 1 .917 0 ~ 0 1 .917 0 places displayed is0 0 0 0 0 0 0 0 somewhatarbitThe general solution is pC = 1.417 pM, pF = .917 pM, with pM free. If pM is assigned the value 100, thenpC = 141.7 and pF = 91.7. Note that only the ratios of the prices are determined. This makes sense, forif the were converted from, say, dollars to yen or Euros, the inputs and outputs of each sector wouldstill balance. The economic equilibrium is not affected by a proportional change in prices. 41. 1.6 Solutions 414. a. Fill in the exchange table one column at a time. The entries in each column must sum to 1.Distribution of Output From :Agric. Energy Manuf . Transp. Purchased by : output input.65 .30 .30 .20 Agric..10 .10 .15 .10 Energy.25 .35 .15 .30 Manuf .0 .25 .40 .40 Transp.b. Denote the total annual output of the sectors by pA, pE, pM, and pT, respectively. From the first row ofthe table, the total input to Agriculture is .65pA + .30pE + .30pM + .20 pT. So the equilibrium pricesmust satisfyincome expensesp A = .65 p A + .30 p E + .30 p M + .20 pTFrom the second, third, and fourth rows of the table, the equilibrium equations arep = .10 p + .10 p + .15 p +.10pp = .25 p + .35 p + .15 p +.30pp = p + p +pE A E M TM A E M TT .25 E .40 M .40TMove all variables to the left side and combine like terms:p p p pp p p pp p p p =.35 .30 .30 .20 0.10 .90 .15 .10 0.25 .35 .85 .30 0A E M TA E M TA E M T + = + =p p p + =.25 .40 .60 0E M TUse gauss, bgauss, and scale operations to reduce the augmented matrix to reduced echelon form.35 .3 .3 .2 0 .35 .3 0 .55 0 .35 0 0 .71 00 .81 .24 .16 0 0 .81 0 .43 0 0 1 0 .53 0 ~ ~ 0 0 1.0 1.17 0 0 0 1 1.17 0 0 0 1 1.17 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0Scale the first row and solve for the basic variables in terms of the free variable pT, and obtainpA = 2.03pT, pE = .53pT, and pM = 1.17pT. The data probably justifies at most two significant figures,so take pT = 100 and round off the other prices to pA = 200, pE = 53, and pM = 120.5. The following vectors list the numbers of atoms of boron (B), sulfur (S), hydrogen (H), and oxygen (O):2 0 1 0 boron3 0 0 1 sulfur B S : , H O: , H BO : , H S:2 3 2 3 3 20 2 3 2 hydrogen0 1 3 0 oxygenThe coefficients in the equation x1B2S3 + x2H20 x3H3BO3 + x4H2S satisfy 42. 42 CHAPTER 1 Linear Equations in Linear Algebra2 0 1 03 0 0 10 2 3 20 1 3 0 + = + x x x x1 2 3 4Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) androw reduce the augmented matrix of the homogeneous system:2 0 1 0 0 2 0 1 0 0 2 0 1 0 0 2 0 1 0 03 0 0 1 0 0 0 3/2 1 0 0 1 3 0 0 0 1 3 0 0 ~ ~ ~ 0 2 3 2 0 0 2 3 2 0 0 0 3/2 1 0 0 0 3/2 1 0 0 1 3 0 0 0 1 3 0 0 0 2 3 2 0 0 0 3 2 0 2 0 1 0 0 2 0 0 2/3 0 1 0 0 1/3 00 1 3 0 0 0 1 0 2 0 0 1 0 2 0~ ~ ~0 0 1 2/3 0 0 0 1 2/3 0 0 0 1 2/3 00 0 3 2 0 0 0 0 0 0 0 0 0 0 0 The general solution is x1 = (1/3) x4, x2 = 2x4, x3 = (2/3) x4, with x4 free. Take x4 = 3. Then x1 = 1,x2 = 6, and x3 = 2. The balanced equation isS3 + 6H20 2H3BO3 + 3H2SB26. The following vectors list the numbers of atoms of sodium (Na), phosphorus (P), oxygen (O),barium (Ba), and nitrogen(N):3 0 0 1 sodium1 0 2 0 phosphorus Na PO : 4 , Ba(NO ) : 6 , Ba (PO ) : 8 , NaNO : 3 oxygen3 4 3 2 3 4 2 30 1 3 0 barium0 2 0 1 nitrogenThe coefficients in the equation x1Na3PO4 + x2Ba(NO3)2 x3Ba3(PO4)2 + x4NaNO3 satisfy3 0 0 11 0 2 04 6 8 30 1 3 00 2 0 1 + = + x x x x1 2 3 4Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) androw reduce the augmented matrix of the homogeneous system: 3 0 0 1 0 1 0 2 0 0 1 0 2 0 0 1 0 2 0 0 1 0 2 0 0 3 0 0 1 0 0 0 6 1 0 0 1 3 0 0 4 6 8 3 0 ~ 4 6 8 3 0 ~ 0 6 0 3 0 ~ 0 6 0 3 0 0 1 3 0 0 0 1 3 0 0 0 1 3 0 0 0 0 6 1 0 0 2 0 1 0 0 2 0 1 0 0 2 0 1 0 0 2 0 1 0 43. 1.6 Solutions 43 1 0 2 0 0 1 0 2 0 0 1 0 0 1/3 00 1 3 0 0 0 1 3 0 0 0 1 0 1/2 0~ 0 0 18 3 0 ~ 0 0 1 1/ 6 0 ~ 0 0 1 1/ 6 00 0 6 1 0 0 0 0 0 0 0 0 0 0 00 0 6 1 0 0 0 0 0 0 0 0 0 0 0The general solution is x1 = (1/3)x4, x2 = (1/2)x4, x3 = (1/6)x4, with x4 free. Take x4 = 6. Then x1 = 2,x2 = 3, and x3 = 1. The balanced equation is2Na3PO4 + 3Ba(NO3)2 Ba3(PO4)2 + 6NaNO37. The following vectors list the numbers of atoms of sodium (Na), hydrogen (H), carbon (C), andoxygen (O):1 0 3 0 0 sodium1 8 5 2 0 hydrogen NaHCO : , H C H O : , Na C H O : , H O: , CO :3 3 6 5 7 3 6 5 7 2 21 6 6 0 1 carbon3 7 7 1 2 oxygenThe order of the various atoms is not important. The list here was selected by writing the elements in theorder in which they first appear in the chemical equation, reading left to right:x1 NaHCO3 + x2 H3C6H5O7 x3 Na3C6H5O7 + x4 H2O + x5 CO2.The coefficients x1, , x5 satisfy the vector equation1 0 3 0 01 8 5 2 01 6 6 0 13 7 7 1 2 + = + + x x x x x1 2 3 4 5Move all the terms to the left side (changing the sign of each entry in the third, fourth, and fifth vectors)and reduce the augmented matrix: 1 0 3 0 0 0 1 0 0 0 1 0 1 8 5 2 0 0 0 1 0 0 1/3 0 ~ ~ 1 6 6 0 1 0 0 0 1 0 1/3 0 3 7 7 1 2 0 0 0 0 1 1 0 The general solution is x1 = x5, x2 = (1/3)x5, x3 = (1/3)x5, x4 = x5, and x5 is free. Take x5 = 3. Then x1 = x4 =3, and x2 = x3 = 1. The balanced equation is3NaHCO3 + H3C6H5O7 Na3C6H5O7 + 3H2O + 3CO28. The following vectors list the numbers of atoms of potassium (K), manganese (Mn), oxygen (O),sulfur (S), and hydrogen (H):1 0 0 0 2 01 1 0 1 0 0KMnO : 4 , MnSO : 4 , H O: 1 , MnO : 2 , K SO : 4 , H SO : 44 4 2 2 2 4 2 40 1 0 0 1 10 0 2 0 0 2potassiummanganeseoxyg ensulfurhydrogenThe coefficients in the chemical equation 44. 44 CHAPTER 1 Linear Equations in Linear Algebrax1KMnO4 + x2MnSO4 + x3H2O x4MnO2 + x5K2SO4 + x6H2SO4satisfy the vector equation1 0 0 0 2 01 1 0 1 0 04 4 1 2 4 40 1 0 0 1 10 0 2 0 0 2 x + x + x = x + x + x 1 2 3 4 5 6 Move the terms to the left side (changing the sign of each entry in the last three vectors) and reduce theaugmented matrix:1 0 0 0 2 0 0 1 0 0 0 0 1.0 01 1 0 1 0 0 0 0 1 0 0 0 1.5 04 4 1 2 4 4 0 ~ 0 0 1 0 0 1.0 00 1 0 0 1 1 0 0 0 0 1 0 2.5 00 0 2 0 0 2 0 0 0 0 0 1 .5 0 The general solution is x1 = x6, x2 = (1.5)x6, x3 = x6, x4 = (2.5)x6, x5 = .5x6, and x6 is free.Take x6 = 2. Then x1 = x3 = 2, and x2 = 3, x4 = 5, and x5 = 1. The balanced equation is2KMnO4 + 3MnSO4 + 2H2O 5MnO2 + K2SO4 + 2H2SO49. [M] Set up vectors that list the atoms per molecule. Using the order lead (Pb), nitrogen (N), chromium(Cr), manganese (Mn), and oxygen (O), the vector equation to be solved is1 0 3 0 0 0 lead6 0 0 0 0 1 nitrogen0 1 0 2 0 0 chromium0 2 0 0 1 0 manganese0 8 4 3 2 1 oxygen + = + + + x x x x x x1 2 3 4 5 6The general solution is x1 = (1/6)x6, x2 = (22/45)x6, x3 = (1/18)x6, x4 = (11/45)x6, x5 = (44/45)x6, andx6 is free. Take x6 = 90. Then x1 = 15, x2 = 44, x3 = 5, x4 = 22, and x5 = 88. The balanced equation is15PbN6 + 44CrMn2O8 5Pb3O4 + 22Cr2O3 + 88MnO2 + 90NO10. [M] Set up vectors that list the atoms per molecule. Using the order manganese (Mn), sulfur (S), arsenic(As), chromium (Cr), oxygen (O), and hydrogen (H), the vector equation to be solved is1 0 0 1 0 0 01 0 1 0 0 3 00 2 0 0 1 0 00 10 0 0 0 1 00 35 4 4 0 12 10 0 2 1 3 0 2 + + = + + + x x x x x x x1 2 3 4 5 6 7manganesesulfurarsenicchromiumoxygenhydrogenIn rational format, the general solution is x1 = (16/327)x7, x2 = (13/327)x7, x3 = (374/327)x7,x4 = (16/327)x7, x5 = (26/327)x7, x6 = (130/327)x7, and x7 is free. Take x7 = 327 to make the othervariables whole numbers. The balanced equation is16MnS + 13As2Cr10O35 + 374H2SO4 16HMnO4 + 26AsH3 + 130CrS3O12 + 327H2O 45. 1.6 Solutions 45Note that some students may use decimal calculation and simply round off the fractions that relate x1,..., x6 to x7. The equations they construct may balance most of the elements but miss an atom or two. Hereis a solution submitted by two of my students:5MnS + 4As2Cr10O35 + 115H2SO4 5HMnO4 + 8AsH3 + 40CrS3O12 + 100H2OEverything balances except the hydrogen. The right side is short 8 hydrogen atoms. Perhaps the studentsthought that the 4H2 (hydrogen gas) escaped!11. Write the equations for each node:Node Flow in Flow outA x 1 x320Bx 2 x 3 x4C 80x xTotal flow: 80 = x 201 24+ == += ++Rearrange the equations:+ = =1 32 3 4+ =1 242008060x xx x xx xx=Reduce the augmented matrix:1 0 1 0 20 1 0 1 0 200 1 1 1 0 0 1 1 0 60 ~ ~ 1 1 0 0 80 0 0 0 1 60 0 0 0 1 60 0 0 0 0 0x3For this type of problem, the best description of the general solution uses the style ofSection 1.2 rather than parametric vector form:x xx xx= 2060is freex 601 32 334= + =. Since x1 cannot be negative, the largest value of x3 is 20.12. Write the equations for each intersection:Intersection Flow in Flow outx x x= + += +A 1 3 440B 200x 1 x2C x + x = x+100D x + x=602 3 54 5Total flow: 200 = 20040200x1 x2x3100x4 x560ABCD2080x1x2x4ACB 46. 46 CHAPTER 1 Linear Equations in Linear AlgebraRearrange the equations: =1 3 41 2+ =+ =2 3 54 54020010060x x xx xx x xx x+ =Reduce the augmented matrix:1 0 1 1 0 40 1 0 1 0 1 1001 1 0 0 0 200 0 1 1 0 1 100 ~ 0 1 1 0 1 100 0 0 0 1 1 60 0 0 0 1 1 60 0 0 0 0 0 0The general solution (written in the style of Section 1.2) isx x xx x xxx xx= + 100100is free60is free1 3 52 3 534 55= + = b. When x4 = 0, x5 must be 60, andx xx xxxx= + 40160is free0601 32 3345= == c. The minimum value of x1 is 40 cars/minute, because x3 cannot be negative.13. Write the equations for each intersection:Intersection Flow in Flow outx + = x+x + x = x +xx + = x+x + = x+x + = x+A 30 80BC 100 40D 40 90E 60 202 13 5 2 46 54 61 3=Total flow: 230 230Rearrange the equations:8060 =1 2 + =2 3 4 55 64 61 330 40AE500605040x xx x x xx xx xx x = = Reduce the augmented matrix:x2CDB 1 1 0 0 0 0 50 1 1 0 0 0 0 50 0 1 1 1 1 0 0 0 1 1 1 1 0 0 0 0 0 0 1 1 60 ~ ~ 0 0 0 1 0 1 50 0 0 0 1 0 1 50 0 0 0 0 1 1 60 1 0 1 0 0 0 40 0 0 0 0 0 0 010090x1 x6x3x5x420 40 47. 1.6 Solutions 47 1 0 1 0 0 0 40 0 1 1 0 0 0 10 ~ ~ 0 0 0 1 0 1 50 0 0 0 0 1 1 60 0 0 0 0 0 0 0a. The general solution isx xx xxx xx xx= 1 32 334 65 664010= + is free5060= + = +is freeb. To find minimum flows, note that since x1 cannot be negative, x340. This implies thatx250. Also, since x6 cannot be negative, x450 and x560. The minimum flows arex2 = 50, x3 = 40, x4 = 50, x5 = 60 (when x1 = 0 and x6 = 0).14. Write the equations for each intersection.Intersection Flow in Flow outx = x+x xx xx xx xx xA 1 2100B 2 + 50=3C 3 = 4+120D 4 + 150=5E 5 = 6+80F + 100=6 1Rearrange the equations:50100 =1 2x3C DB EA =2 3 =3 4 = 4 5 =5 61 6120 150x21005012015080100x xx xx xx xx xx x + = Reduce the augmented matrix:80100x5 x4x1x6F 1 1 0 0 0 0 100 1 1 0 0 0 0 100 0 1 1 0 0 0 50 0 1 1 0 0 0 50 0 0 1 1 0 0 120 0 0 1 1 0 0 120 ~ ~ 0 0 0 1 1 0 150 0 0 0 1 1 0 150 0 0 0 0 1 1 80 0 0 0 0 1 1 80 1 0 0 0 0 1 100 0 0 0 0 0 0 0 48. 48 CHAPTER 1 Linear Equations in Linear Algebra 1 0 0 0 0 1 1000 1 0 0 0 1 00 0 1 0 0 1 50 ~ ~0 0 0 1 0 1 700 0 0 0 1 1 800 0 0 0 0 0 0 . The general solution isx xx xx xx xx xx= 100+ 1 62 63 64 65 66= = + 507080is free= + = +.Since x4 cannot be negative, the minimum value of x6 is 70.Note: The MATLAB box in the Study Guide discusses rational calculations, needed for balancing thechemical equations in Exercises 9 and 10. As usual, the appendices cover this material for Maple,Mathematica, and the TI and HP graphic calculators.1.7 SOLUTIONSNote: Key exercises are 920 and 2330. Exercise 30 states a result that could be a theorem in the text. Thereis a danger, however, that students will memorize the result without understanding the proof, and then latermix up the words row and column. Exercises 37 and 38 anticipate the discussion in Section 1.9 of one-to-onetransformations. Exercise 44 is fairly difficult for my students.1. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are thethree given vectors. Since5 7 9 0 5 7 9 00 2 4 0 ~ 0 2 4 00 6 8 0 0 0 4 0 , there are no free variables. So thehomogeneous equation (*) has only the trivial solution. The vectors are linearly independent.2. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are thethree given vectors. Since 0 0 3 0 2 8 1 0 0 5 4 0 ~ 0 5 4 0 2 8 1 0 0 0 3 0, there are no free variables. So thehomogeneous equation (*) has only the trivial solution. The vectors |
Linear algebra and its applications 3rd edition solutions
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