(Solution Manual) Linear Algebra and its Applications

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Solution Manual

3rd Edition

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Solution Manual for Linear Algebra and its Applications, The downloadable file consists of 423 pages, and the total size of the file is 2.78 MB. Purchase now only for highly discounted price.

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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+= −− = − 157 275 −−− Replace R2 by R2 + (2)R1 and obtain: 12 257 39xx x+= = 157 039     Scale R2 by 1/3: 12 257 3xx x+= = 157 013     Replace R1 by R1 + (–5)R2: 1 28 3x x=− = 10 8 01 3−      The solution is ( x1, x2) = (–8, 3), or simply (–8, 3). 2. 12 1224 4 57 1 1xx xx+= − += 24 4 571 1−      Scale R1 by 1/2 and obtain: 12 1222 57 1 1xx xx+= − += 12 2 571 1−      Replace R2 by R2 + (–5)R1: 12 222 32 1xx x+= − −= 122 03 2 1−    −   Scale R2 by –1/3: 12 222 7xx x+= − =− 12 2 01 7−    −   Replace R1 by R1 + (–2)R2: 1 212 7x x= =− 10 1 2 01 7   −   The solution is ( x1, x2) = (12, –7), or simply (12, –7). 2 CHAPTER 1 • Linear Equations in Linear Algebra 3. The point of intersection satisfies the system of two linear equations: 12 1257 22xx xx+= −= − 157 122   −−   Replace R2 by R2 + (–1)R1 and obtain: 12 257 79xx x+= −= − 157 079   −−   Scale R2 by –1/7: 12 257 9/7xx x+= = 15 7 019 / 7     Replace R1 by R1 + (–5)R2: 1 24/7 9/7x x= = 104 / 7 01 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: 12 1251 37 5xx xx−= −= 15 1 37 5−  −  Replace R2 by R2 + (–3)R1 and obtain: 12 251 82xx x−= = 15 1 08 2−      Scale R2 by 1/8: 12 251 1/4xx x−= = 15 1 01 1 /4−      Replace R1 by R1 + (5)R2: 1 29/4 1/4x x= = 109 / 4 011 / 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 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 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, which produces 16401 02704 00123 0005 1 5−−  −  − − . 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 are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty. 1.1 • Solutions 3 8. The standard row operations are: 1 490 1 490 1 400 1000 01 7 0 ~ 01 7 0 ~ 01 0 0 ~ 0 1 0 0 00 2 0 00 1 0 00 1 0 0 0 1 0−−−                         The solution set contains one solution: (0, 0, 0). 9. The system has already been reduced to triangular fo rm. Begin by scaling the fourth row by 1/2 and then replacing R3 by R3 + (3)R4: 11004 11004 110 04 01307 01307 013 07~~0013 1 00 131 001 05 00024 00012 000 12−− −− − −        −− − − −        −− −−         Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2: 1 100 4 10004 01 0 08 0 1 0 0 8~~00 1 05 0 0 1 0 5 0 001 2 00012−−                   The 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 the –4 and 3 above it to zeros. That is, replace R2 by R2 + (4)R4 and replace R1 by R1 + (–3)R4. For the final step, replace R1 by R1 + (2)R2. 1 20 3 2 1 200 7 1000 3 01 047 01 0 05 0 1 0 05~~00 106 00 1 06 0 0 1 06 0 00 1 3 0 001 3 0001 3−− − −          −− −              −− −      The 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 45 1 3 52 1352 1 3 52 1 3 5 2~0 1 4 5~0 1 4 5~0 1 4 5 3 7 76 3 7 76 028 1 2 0 0 02−− −−          −− −−          −−      The 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. 1344 1344 1344 3778 ~ 0254 ~ 0254 4617 06 1 59 0003−− −− −−          −− − −          −− −−     The system is inconsistent, because the last row would require that 0 = 3 if there were a solution. The solution set is empty. 4 CHAPTER 1 • Linear Equations in Linear Algebra 13. 1 038 1 038 1 038 1 038 2 297 ~ 0 2 1 59 ~ 0 152 ~ 0 152 0 152 0 152 0 2 1 59 0 055−−−−        −−−        −−−−     10 3 8 100 5 ~0 1 5 2~0 1 0 3 00 1 1 001 1−      −      −−   . The solution is (5, 3, –1). 14. 1 305 1 305 1 305 1 305 1 1 5 2~0 2 5 7~0 1 1 0~0 1 1 0 0 110 0 110 0 257 0 077−−−−          −−          −      13 0 5 13 05 1 0 02 ~0 1 1 0~0 1 0 1~0 1 0 1 . 0 011 0 01 1 001 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. 10 302 1030 2 01 03301 03 3~02 3210232 1 30 075 00971 1     −−      −−   −− −    10 3 0 2 103 0 2 01 0 3 3 01 0 3 3~~00 3 4 7 003 4 7 00 9 7 1 1 000 51 0     −−      −−   −− −    The resulting triangular system i ndicates that a solution exists. In fa ct, 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 before adding to R4, but the arithmetic is rather easy keep ing R2 unchanged.) Finally, replace R4 by R4 + R3. 1 0 023 1 0 023 022 0 0 022 0 0~001 3 1 001 3 1 232 1 5 032 3 1−− −−               −− −    1 0023 1 0 023 02 2 0 0 022 0 0~~00 1 3 1 001 3 1 00 1 3 1 000 0 0−− −−               −−−    The system is now in triangular form and has a solu tion. The next section discusses how to continue with this type of system. 1.1 • Solutions 5 17. Row reduce the augmented matrix corresponding to the given system of three equations: 141 14 1 14 1 21 3 ~ 075 ~ 075 134 075 000−−−          − −−−          −− −     The system is consistent, and using the argument fro m Example 2, there is only one solution. So the three lines have only one point in common. 18. Row reduce the augmented matrix corresponding to the given system of three equations: 12 14 12 1 4 12 1 4 01 11 ~ 01 1 1 ~ 01 1 1 13 00 01 1 4 00 0 5         −− −          −− −      The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point in common. 19. 14 1 4~368 06 3 4hh h     −−    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. 13 1 3~.24 6 04 2 0hh h−−      −+   Write c for 4 + 2 h. Then the second equation cx2 = 0 has a solution for every value of c. So the system is consistent for all h. 21. 13 2 1 3 2~.48 0 1 2 0 hh−−      −+   Write c for h + 12. Then the second equation cx2 = 0 has a solution for every value of c. So the system is consistent for all h. 22. 23 23~.69 5 00 5 3hh h−−      −+   The system is consistent if and only if 5 + 3 h = 0, that is, if and only if 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 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. 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 that transforms 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 paragr aph after equation (2). 6 CHAPTER 1 • Linear Equations in Linear Algebra 25. 147 147 147 035 ~ 035 ~ 035 259 035 2 000 2gg g hh h kk gk gh−− −          −− −          −− − + + +      Let b denote the number k + 2g + h. Then the third equation represente d 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. A basic principle of this section is that row opera tions 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 perform any elementary row operations to produce other augm ented matrices. Here are three examples. The fact that they are all row equivalent proves that they all have the solution set (–2, 1, 0). 100 2 100 2 100 2 01 0 1 ~ 21 0 3 ~ 21 0 3 001 0 001 0 201 4−−−      −−      −    27. Study the augmented matrix for the given system, replacing R2 by R2 + (– c)R1: 13 1 3~03f f cd g d cg c f     −−    This shows that shows d – 3c must be nonzero, since f and g are arbitrary. 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 ≠ 3c. 28. 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 a         −−      The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be written as ad – bc ≠ 0, or ad ≠ bc. 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: 21 3 2 1 3(2 0 4 0) /4,o r 4 6 0 TT T T T T=++ + − − = 34 2 3 4 2( 40 30)/4, or 4 70 TT T T T T=+ + + − − = 4 13 413(10 30)/4, or 4 40 TT T T TT =+ + + − − = 1.1 • Solutions 7 Rearranging, 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 1 014 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 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: 10 144 0 1 0 144 0 1 0 144 0 01 01 50 1 01 50 1 01 5~~ ~01417 0 0 0427 5 0 0 427 5 0 1 4 15 190 0 0 4 14 195 0 0 0 12 270−− −− −−          −− −          −− − −     −− −      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 0 1 0 1 5 0 1 0 0 27.5 0 1 0 0 27.5~~ ~004 2 7 5 0040 1 2 0 0010 3 0 0 0 0 1 22.5 0 0 0 1 22.5 0 0 0 1 22.5−−          −          −           The last step is to replace R1 by R1 + (–1)R3: 10002 0 . 0 0 1 0 0 27.5~.00103 0 . 0 0 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, even the small grid for Exercise 34 leads to about 25 multip lications or additions (not counting operations with zero). This exercise should give students an apprecia tion for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states th e solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses th is 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, Mathemat ica, and the TI-83+/86/89 and HP-48G 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 a ll 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 fr om the text web site, www.laylinalgebra.com . 8 CHAPTER 1 • Linear Equations in Linear Algebra 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. 1234 1 2 3 4 1 2 3 4 4567 ~ 0 3 6 9 ~ 0 1 2 3 6 7 8 9 051 01 5 051 01 5     −−−      −− − −− −    1234 1 0 1 2 ~0 1 2 3~0 1 2 3 0000 0 0 0 0−−               . Pivot cols 1 and 2. 1234 4567 6789          4. 1357 1 3 5 7 1 3 5 7 1 3 5 7 3 5 7 9~0 4 8 1 2~0 1 2 3 ~0 1 2 3 5 7 9 1 0 8 16 34 0 8 16 34 0 0 0 10             −−−              −− − −− − −        1357 1350 10 10 ~0 1 2 3~0 1 2 0~0 1 2 0 0001 0001 00 01−                    . Pivot cols 1, 2, and 4 1357 3579 5791          5. ** 0,,00 000          „„ „ „ 6.** 0 0, 0 0 ,0 0 00 00 00                  „„ „ „ 7. 1347 13 4 7 1347 130 5~~ ~3976 00 5 1 5 00 13 00 1 3−            −−       Corresponding system of equations: 12 335 3xx x+= − = 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 is 12 2 353 is free 3x x x x=− −   =  Note : Exercise 7 is paired with Exercise 10. 1.2 • Solutions 9 8. 1 4 0 7 1 4 0 7 1407 100 9~~ ~2 7 0 1 00 1 0 40 1 0 40 1 0 4−              −−        Corresponding system of equations: 1 29 4x x=− = 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 vari able. In this particular problem, the basic variables do not depend on the value of the free variable. General solution: 1 2 39 4 is freex x x=−  =   Note : A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.) 9. 01 65 1 276 1 05 4~~1276 0165 0 16 5−− − −        −− − −     Corresponding system: 13 2354 65xx xx−= −= Basic variables: x1, x2; free variable: x3. General solution: 13 23 345 56 is freex x x x x=+  =+   10. 12 1 3 1213 12 04~~362 2 0017 00 17−− −− − −          −− − −      Corresponding system: 12 324 7xx x−= − =− Basic variables: x1, x3; free variable: x2. General solution: 12 2 342 is free 7x x x x=− +   =−  11. 342 0 3420 14 /3 2 /3 0 9 1 26 0 ~ 0000 ~ 0 0 0 0 684 0 0000 0 0 0 0−− −          −−          −−     Corresponding system: 12342033 00 00xx x−+= = = 10 CHAPTER 1 • Linear Equations in Linear Algebra Basic variable: x1; free variables x2, x3. General solution: 123 2 342 33 is free is freex x x x x=−      12. 17065 17065 17 065 00123 ~ 00123 ~ 00 123 17427 0048 1 2 00 000−−−          −− −− −−          −− −     Corresponding system: 12 4 3476 5 23 00xx x xx−+ = −= − = Basic variables: x1 and x3; free variables: x2, x4. General solution: 12 4 2 34 457 6 is free 32 is freex xx x xx x=+ −  =− +  13. 1 3 0 1 0 2 1 300 9 2 1000 35 0 1 0 0 4 1 0 100 4 1 0100 41~~000194 00 0 19 4 0 0 0 19 4 0 0 0 0 0 0 0 000 0 0 0000 00−−− − −          −− −                    Corresponding system: 15 25 4535 41 94 00xx xx xx−= −= += = Basic variables: x1, x2, x4; free variables: x3, x5. General solution: 15 25 3 45 553 14 is free 49 is freex x x x x x x x=+  =+ =−  Note : The Study Guide discusses the common mistake x3 = 0. 14. 1 256 05 1 070 09 0 163 02 0 163 02~0 000 10 0 000 10 0 000 00 0 000 00−− − −    −− −−        1.2 • Solutions 11 Corresponding system: 13 23 4 579 63 2 0 00xx xxx x+= − −− = = = Basic variables: x1, x2, x5; free variables: x3, x4. General solution: 13 23 4 3 4 597 26 3 is free is free 0xx x xx x x x=− −  =+ +   =  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. 23 23~467 007 2hh h     −    The system has a solution only if 7 – 2 h = 0, that is, if h = 7/2. 18. 132 1 3 2~57 0 1 5 3 hh−− − −      −+    If h +15 is zero, that is, if h = –15, then the system has no solution, because 0 cannot equal 3. Otherwise, when 15, h≠− the system has a solution. 19. 12 1 2~48 08 4 8hh kh k     −−    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. 1 3 213 2~30 9 6 hk h k     −−    a. When h = 9 and 6, k≠ the system is inconsistent, because th e augmented column is a pivot column. b. When 9,h≠ 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 5 x4 = 0, which does not by itself lead to a contradiction. So the system might be c onsistent or it might be inconsistent. 12 CHAPTER 1 • Linear Equations in Linear Algebra 22. a. False. See the statement pr eceding 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 are determined completely by the positions of the l eading entries in the nonzero rows of any echelon form 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 pi vots, there must be a pivot in the third (bottom) row of the coefficient matrix. The reduced eche lon 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 ther e 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 system has no solution. 25. If the coefficient matrix has a pivot position in ever y row, then there is a pivot position in the bottom row, and there is no room for a pivot in the augm ented column. So, the system is consistent, by Theorem 2. 26. Since there are three pivots (one in each row), the augmented matrix must reduce to the form 1 2 3100 0 1 0 and so 001ax a bx b cx c=  =  =  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 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 variab les 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 sy stem is consistent, each different value of a free variable will produce a different solution. 30. Example: 123 1234 2225xxx xxx++= ++= 31. Yes, a system of linear equations with more equations than unknowns can be consistent. Example (in which x1 = x2 = 1): 12 12 122 0 32 5xx xx xx+= −= += 1.2 • Solutions 13 32. According to the numerical note in Section 1.2, when n = 30 the reduction to echelon form takes about 2(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 fracti on associated with the backward phase is about (9×104) /(18×106) = .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), the coefficients a0, a1, a2 must satisfy the systems of equations gi ven in the text. Row reduce the augmented matrix: 111 1 2 111 1 2 1111 2 1111 2 1241 5 ~ 0 13 3 ~ 0 13 3 ~ 0 13 3 1391 6 028 4 002 2 001 1                            −−        1101 3 100 7 ~0 1 0 6~0 1 0 6 001 1 001 1        −−   The polynomial is p(t) = 7 + 6 t – t2. 34. [M] The system of equations to be solved is: 2345 01 2 3 4 5 2345 0 1 2345 2345 0 1 2345 2345 01 2 3 4 5 2345 01 2 3 4 5 23 01 2 30 0000 0 22 2 2 2 2 .90 44 4 4 4 1 4.8 6 6666 39.6 88 8 8 8 7 4.3 10 10 10a a aaaa aa a a a a aa a a a a a a aaaa aa a a a a aa a a+⋅ +⋅+⋅+⋅+⋅= +⋅ + ⋅ +⋅+⋅ +⋅= +⋅ + ⋅ +⋅+⋅ +⋅= +⋅ +⋅+⋅+⋅+⋅= +⋅ +⋅+⋅+⋅+⋅= +⋅+⋅ +⋅ +45 4510 10 119aa⋅+ ⋅= The unknowns are a0, a1, …, a5. 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 1 2 4 8 16 32 2.9 0 2 4 8 16 32 2.9 1 4 16 64 256 1024 14.8 0 0 8 48 224 960 9~1 6 36 216 1296 7776 39.6 0 0 24 192 1248 7680 30.9 1 8 64 512 4096 32768 74.3 0 0 48 480 4032 32640 62.7 0 0 80 960 9920 99840 10 11 01 0 1 0 1 0 1 0 1 1 9        4.5                   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 0 0 8 48 224 960 9 0 0 8 48 224 960 9~~000 4 8 5 7 6 4 8 0 0 3 . 9 0004 8 5 7 6 4 8 0 0 3 . 9 0001 9 22 6 8 82 6 8 8 0 8 . 7 000 0 3 8 4 7 6 8 0 6 . 9 0004 8 07 6 8 09 0 2 4 01 4 . 5 000 01 9 2 04 2 2 4 0 2 4 . 5              − −      14 CHAPTER 1 • Linear Equations in Linear Algebra 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 64 8 0 0 3 . 9 0004 85 7 64 8 0 0 3 . 9 0 0 0 0 384 7680 6.9 0 0 0 0 384 7680 6.9 0 0 0 0 0 3840 10 0 0 0 0 0 1 .0026                     −−         100 0 00 0 100000 0 024 8 1 60 2 . 8 1 6 7 010000 1 . 7 1 2 5 0 0 8 48 224 0 6.5000 0 0 1 0 0 0 1.1948~~ ~0004 85 7 60 8 . 6 0 0 0 000100 . 6 6 1 5 0 0 0 0 384 0 26.900 0 0 0 0 1 0 .0701 000 0 01. 0 0 2 6 0 4 000001 . 0 0 2 6            −   −      −−        " Thus p(t) = 1.7125 t – 1.1948 t2 + .6615 t3 – .0701 t4 + .0026 t5, and p(7.5) = 64.6 hundred lb. Notes : In Exercise 34, if the coefficients are reta ined to higher accuracy than shown here, then p(7.5) = 64.8. If a polynomial of lower degree is used, the resulting sy stem 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 phras e, “If and only if,” used in Theorem 2. 1.3 SOLUTIONS Notes : The key exercises are 11–14, 17–22, 25, and 26. A discussion of Exercise 25 will help students understand the notation [ a1 a2 a3], {a1, a2, a3}, and Span{ a1, a2, a3}. 1. 131 ( 3) 4 212 ( 1)1−−− +− −  += + = =  −+ −  uv . Using the definitions carefully, 13 1 (2)(3) 16 52( 2)21 2 (2)(1) 22 4−− − −− − +     −= + − = + = =     −− − +     uv , or, more quickly, 13 1 6 52221 2 2 4−− − +    −= − = =    −+    uv . The intermediate step is often not written. 2. 32 3 2 5 21 2 ( 1) 1+      += + = =      −+ −      uv . Using the definitions carefully, 1.3 • Solutions 15 32 3 (2)(2) 3(4) 12( 2)2 12( 2 )( 1) 2 2 4−+ − −      −= + − = + = =      −− − +      uv , or, more quickly, 32 3 4 12221 2 2 4−−    −= − = =    −+    uv . The intermediate step is often not written. 3. x2 x1u – 2v – 2vu – v – v vu u + v 4. x2 x1u – v u vu + v – v– 2vu – 2v 5. 1263 1 14 7 50 5xx−    −+ = −    −  , 12 12 163 1 47 50 5xx xx x−        −+ = −        −    , 12 12 163 1 47 55xx xx x−      −+ = −     −   12 12 163 1 47 55xx xx x−= −+ = − =− Usually the intermediate steps are not displayed. 6. 12 328 1 0 35 6 0xx x−    ++ =     −     , 3 12 3 1228 0 6 35 0x xx x xx−     ++ =     −     , 12 3 12328 0 35 6 0xx x xx x−+ + =  +−   223 12328 0 35 60xxx xx x−+ + = +−= Usually the intermediate steps are not displayed. 7. See the figure below. Since the grid can be extende d 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. 16 CHAPTER 1 • Linear Equations in Linear Algebra 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 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) – 1.5 v = 2u – 3.5 v 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 = –3 v + 3u – v = 3u – 4v Another solution is d = b – 2v + u = (2 u – 2v) – 2v + u = 3u – 4v w xvu acd 2vb zy –2v–u–v0 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 = –2 u + 2v. Or, use the fact that x is –1 units in the u-direction from w, so that x = w – u = (– u + 2v) – u = –2 u + 2v y. The vector y is 1.5 units from x in the v-direction, so y = x + 1.5 v = (–2 u + 2v) + 1.5 v = –2 u + 3.5 v 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 = –2 u + 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+    +− =    −+ −   23 12 3 12 305 0 46 0 38 0xx xx x xx x           ++ − =           −−     , 12 301 5 0 46 1 0 13 8 0xx x         ++ − =         −−     Usually, the intermediate calculations are not displayed. 1.3 • Solutions 17 Note : The Study Guide says, “Check with your instructor whether you need to “show work” on a problem such as Exercise 9.” 10. 123 123 12343 9 72 2 86 51 5xx x xx x xx x++= −−= +−= , 12 3 123 12343 9 72 2 865 1 5xx x xxx xxx++    −− =    +−   123 123 12343 9 72 2 86 5 1 5xxx xxx xxx         +− +− =          −      , 12 341 3 9 172 2 865 15xx x         +− +− =          −      Usually, the intermediate calculations are not displayed. 11. The question Is b a linear combination of a1, a2, and a3? is equivalent to the question Does the vector equation x1a1 + x2a2 + x3a3 = b have a solution? The equation 12 3 12310 5 2 21 6 1 02 8 6xx x       −+ + −= −           ↑↑↑ ↑ aa a b (*) has the same solution set as the lin ear system whose augmented matrix is 10 5 2 21 6 1 02 8 6M =− − −  Row reduce M until the pivot positions are visible: 1052 1052 ~0 1 4 3~0 1 4 3 0286 0000M          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 a3. 12. The equation 12 3 12 3102 5 250 1 1 258 7xx x−          −+ + =          −      ↑↑ ↑ ↑ aa a b (*) has the same solution set as the lin ear system whose augmented matrix is 18 CHAPTER 1 • Linear Equations in Linear Algebra 102 5 250 1 1 258 7M−  =−  −  Row reduce M until the pivot positions are visible: 102 5 102 5 ~0 5 4 1~0 5 4 1 054 3 000 2M−−           The linear system corresponding to M has no solution, so the vector equation (*) has no solution, and therefore 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 the boxed fact on page 34. Row reduced the au gmented matrix until you reach echelon form: 142 3 14 23 0357 ~ 03 57 2843 00 03−−      −−      −− −    The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns of A. 14. [a1 a2 a3 b] = 126 1 1 126 1 1 0375 ~ 0375 1259 00 1 12−− −−    −−    −−  . The linear system corresponding to this matrix 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 0· v1 + 0·v2 = 0, and 1· v1 + 0·v2 = 7 1 6   −, 0·v1 + 1·v2 = 5 3 0−          1· v1 + 1·v2 = 2 4 6   −, 1·v1 – 1·v2 = 12 2 6   −   −  16. Some likely choices are 0· v1 + 0·v2 = 0, and 1· v1 + 0·v2 = 3 0 2   , 0·v1 + 1·v2 = 2 0 3−    1· v1 + 1·v2 = 1 0 5   , 1·v1 – 1·v2 = 5 0 1   − 1.3 • Solutions 19 17. [a1 a2 b] = 12 4 12 4 12 4 12 4 43 1 ~ 05 1 5 ~ 01 3 ~ 01 3 27 03 8 03 8 00 1 7 hh h h−− − −            −− − −            −+ + +      . The vector b is in Span{ a1, a2} when h + 17 is zero, that is, when h = –17. 18. [v1 v2 y] = 13 13 13 015 ~ 01 5~ 01 5 283 023 2 00 7 2hh h hh−− −          −− −          −− − + +     . The vector y is in Span{ v1, v2} when 7 + 2 h 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. For instance, a v1 + bv2 = av1 + b(3/2) v2 = (a + 3b/2)v1 So 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 multiple of the other. Every vector in the set has 0 as its second entry and so lies in the xz-plane in ordinary 3-space. So Span{ v1, v2} is the xz-plane. 21. Let y = h k . Then [ u v y] = 22 22~11 02 / 2hh kk h     −+  . 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 corre sponds to an inconsistent system. Perform sufficient row operations on the matrix to eliminate a ll zero entries in the first three columns. 23. a. False. The alternative notation for a (column) v ector 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−  were on the line through 2 5−  and the origin, then 5 2−     would have to be a multiple of 2 5− , 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). 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. True. See the beginning of the subsection Vectors in Rn. b. True. Use Fig. 7 to draw th e 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 CHAPTER 1 • Linear Equations in Linear Algebra 25. 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 method of Exercise 13. 12 310 4 4 10 44 10 44 03 2 1 ~ 03 21 ~ 03 21 26 3 4 06 54 00 12−− −        −− −        −− − −    ↑↑ ↑ ↑ aa a b The 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] = 20 6 1 0 10 3 5 1035 1 0 3 5 18 5 3 ~ 18 5 3 ~ 0888 ~ 0 8 8 8 1 21 3 1 213 0 2 2 2 0000−− −− − − −                                   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 = 0·a1 + 0·a2 + 1·a3. 27. a. 5v1 is the output of 5 days’ operation of mine #1. b. The total output is x1v1 + x2v2, so x1 and x2 should satisfy 11 2 2150 2825xx +=   vv . c. [M] Reduce the augmented matrix 20 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 of bituminous coal is 27.6 x1 + 30.2 x2 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 360xx + . c. [M] The appropriate values for x1 and x2 satisfy 1227.6 30.2 162 3100 6400 23,610 250 360 1,623xx       +=           . To solve, row reduce the augmented matrix: 27.6 30.2 162 1.000 0 3.900 3100 6400 23610 ~ 0 1.000 1.800 250 360 1623 0 0 0          The steam plant burned 3.9 tons of anthrac ite coal and 1.8 tons of bituminous coal. 1.3 • Solutions 21 29. The total mass is 2 + 5 + 2 + 1 = 10. So v = (2 v1 +5v2 + 2v3 + v4)/10. That is, 5 4 4 9 10 20 8 9 1.3 1124 53 23 8 8 1 5 6 8 . 910 103 2 1 6 61 026 0 −− + − −            =− ++ −+= − +−+=           −− − −+      v 30. Let m be the total mass of the system. By definition, 1 11 11()k kk km mmmmm m=+ + = + + vv v v v "" The second expression displays v as a linear combination of v1, …, vk, which shows that v is in Span{ v1, …, vk}. 31. a. The center of mass is 082 1 0/3 1111114 2 3    ⋅+ ⋅+ ⋅ =        . b. The total mass of the new system is 9 grams. The three masses added, w1, w2, and w3, satisfy the equation () () ()123082 2 1111114 2 9www    +⋅ + +⋅ + +⋅ =          which can be rearranged to () ( ) () 123082 18111114 18www    +⋅ + +⋅ + +⋅ =          and 123082 8 114 12www    ⋅+ ⋅+ ⋅=         The condition w1 + w2 + w3 = 6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown be low, along with a sequence of row operations: 111 6 1116 1116 082 8 ~ 0828 ~ 0828 1 1 4 1 20 0 3 60 0 1 2                         1104 1003 . 5 1003 . 5 ~0 8 0 4~0 8 0 4~0 1 0 . 5 0012 001 2 001 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 of mass 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. The darker parallelogram shows that b is a linear combination of v1 and v2, that is c 1v1 + c2v2 + 0·v3 = b 22 CHAPTER 1 • Linear Equations in Linear Algebra The larger parallelogram shows that b is a linear combination of v1 and v3, that is, c 4v1 + 0·v2 + c3v3 = b So the equation x1v1 + x2v2 + x3v3 = b has at least two solutions, not just one solution. (In fact, the equation has infinitely many solutions.) c2v2c3v3 0v3 c4v1c1v1v1v2b 33. a. For j = 1,…, n, the jth entry of ( u + v) + w is (uj + v j) + wj. By associativity of addition in R, this entry equals uj + (vj + w j), 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 + v j), and the jth entry of cu + cv is cuj + cv j (by definition of scalar multiplication and vector additio n). These entries are equal, by a distributive law in R. So c(u + v ) = cu + cv. 34. a. For j = 1,…, n, uj + (–1)uj = (–1)uj + u j = 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. These entries in R are equal, so the vectors c(du) and ( cd)u are equal. Note : When an exercise in this section involves a vect or equation, the correspondi ng technology data (in the data files on the web) is usually pres ented as a set of (column) vectors. To use MATLAB or other technology, a student must first construct an augmented matr ix from these vectors. The MATLAB note in the Study Guide describes how to do this. The appendices in the Study Guide give corresponding information about Maple, Mathematica, and the TI and HP calculators. 1.4 SOLUTIONS Notes : Key exercises are 1–20, 27, 28, 31 and 32. Exercises 29, 30, 33, and 34 are harder. Exercise 34 anticipates 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 3×2 matrix 42 16 01−    does not match the number of entries (3) in the vector 3 2 7   −     . 1.4 • Solutions 23 2. The matrix-vector product Ax product is not defined because the number of columns (1) in the 3×1 matrix 2 6 1   − does not match the number of entries (2) in the vector 5 1 −. 3. 65 6 5 1 2 1 5 3243 2 4 3 3 8 9 1376 7 6 1 4 1 8 4A−−                =− − = − − − =− + =          −        −−        x , and 65 6 2 5 ( 3 ) 324 3 (4 )2 (3 )(3 ) 1376 7 2 6 ( 3 ) 4A⋅+⋅ − −          =− − = − ⋅ +− ⋅ − =       −     ⋅+⋅ − −     x 4. 183 4 8 3 4 8 3 4 711 1 151 2 5 1 2 5 1 2 81A−− +−         == ⋅ +⋅ +⋅ = =          ++        x , and 183 4 8 1 3 1 ( 4 ) 1 7151 2 5 1 1 1 2 1 81A−⋅ +⋅+−⋅     == =     ⋅+⋅+ ⋅    x 5. On the left side of the matrix e quation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A: 5184 8513227 3 5 1 6−−    ⋅− ⋅+ ⋅− ⋅=    −− −    6. On the left side of the matrix e quation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A: 73 1 21 92596 1 2 32 4−    − −⋅ − ⋅ =  −  −−  7. The left side of the equation is a linear co mbination of three vectors. Write the matrix A whose columns are those three vectors, and create a variable vector x with three entries: 457 4 5 7 138 1 3 8 750 7 5 0 41 2 4 1 2A −−      −− − −  ==   −−   −−   and 1 2 3x x x = x . Thus the equation Ax = b is 1 2 3457 6 138 8 750 0 41 2 7x x x−      −− −     =    −    −−   24 CHAPTER 1 • Linear Equations in Linear Algebra For your information : The unique solution of this equation is (5, 7, 3). Finding the solution by hand would be time-consuming. Note : The skill of writing a vector equation as a matrix equation will be important for both theory and application throughout the text. See also Exercises 27 and 28. 8. The left side of the equation is a linear co mbination of four vectors. Write the matrix A whose columns are those four vectors, and create a variable vector with four entries: 4 4 5 3 445 3 254 0 2 5 4 0A −− − −    ==   −−   , and 1 2 3 4z z z z      =      z . Then the equation Az = b is 1 2 3 4445 3 4 254 0 1 3z z z z −−   =   − . For your information : One solution is (7, 3, 3, 1). The general solution is z1 = 6 + .75 z3 – 1.25 z4, z2 = 5 – .5 z3 – .5z4, with z3 and z4 free. 9. The system has the same solution set as the vector equation 12 331 5 9 01 4 0xxx−      ++ =            and this equation has the same solution set as the matrix equation 1 2 331 5 9 01 4 0x x x−    =       10. The system has the same solution set as the vector equation 1281 4 54 1 13 2xx−        +=        −     and this equation has the same solution set as the matrix equation 1 281 4 54 1 13 2x x−      =       −    11. To solve Ax = b, row reduce the augmented matrix [ a1 a2 a3 b] for the corresponding linear system: 1242 1 2 42 1 2 42 1 2 06 1 0 00 0 1 5 2~0 1 5 2~0 1 5 2~0 1 0 3~0 1 0 3 243 90 0 550 0 110 0 110 0 1 1−−−−            −−            −−−      1.4 • Solutions 25 The solution is 1 2 30 3 1x x x=  =−  = . As a vector, the solution is x = 1 2 30 3 1x x x     =−        . 12. To solve Ax = b, row reduce the augmented matrix [ a1 a2 a3 b] for the corresponding linear system: 12 10 1 2 10 1 2 10 1 2 1 0 3 1 2 1~0 5 5 1~0 5 5 1~0 5 5 1 05 31 0 5 31 0 022 0 0 1 1             −−              −− − −        120 1 120 1 100 3 / 5 ~0 5 0 4~0 1 0 4 / 5~0 1 0 4 / 5 0 0 1 10 0 11 0 0 11−−          −− −               The solution is 1 2 33/5 4/5 1x x x=  =−  = . As a vector, the solution is x = 1 2 33/5 4/5 1x x x     =−        . 13. The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of the columns of A. This happens if and only if the equation Ax = u has a solution. (See the box preceding Example 3 in Section 1.4.) To study this equation, reduce the augmented matrix [ A u] 35 0 11 4 11 4 1 1 4 26 4 ~ 26 4 ~ 081 2 ~ 0 8 1 2 11 4 35 0 081 2 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: 5 872 1 302 1302 1 30 2 0 113 ~ 0 113 ~ 0 113 ~ 0 11 3 1 302 5 872 0778 0 002 9           −− −− −− − −            −− −       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 1 221 63b b−    − , which is row equivalent to 1 2121 00 3b bb−  + . This shows that the equation Ax = b is not consistent when 3 b1 + b2 is nonzero. The set of b for which the equation is consistent is a line through the origin–the set of all points ( b1, b2) satisfying b2 = –3 b1. 16. Row reduce the augmented matrix [ A b]: 1 2 3134 326 , . 518b Ab b−−    =− =    −−  b 11 22 1 33 1134 134 326 ~ 076 3 5 1 8 0 14 12 5bb bb b bb b−− −−      −− − +      −− −   