Linear Algebra and Its Applications习题解答

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Linear Algebra and Its Applications习题解答
1.1 Solution 8. The standard row operations are 901「1-4901「1-400 0170-0170-0100-0100 The 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 then replacing r3 by R3+(3)R4 01-30-701-30701-30-7 00105 2|00012 Next, replace R2 by R2+(3R3. Finally, replace Rl by rl+ R2 0004 1000 1-100-4 010080100 01050010 0001 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 rl by r1+(3R4 For the final step, replace rl by r1 +(2)R2 1-203-2「1-20071「1000 00106 000 06001 -3000 001 The solution set contains one solution: (3,-5,6,-3). 11. First, swap RI and r2 Then replace R3 by R3+(3Rl. Finally, replace R3 by R3+(2)R2 135-2 8120002 The system is inconsistent. because the last row would require that o=2 if there were a solution The solution set Is empty 12. Replace R2 by r2+(3RI and replace R3 by r3+(4Rl. Finally, replace R3 by R3+(3)R2 1-34 46-170-615 0003 The system is inconsistent, because the last row would require that=3 if there were a solution The solution set is empty. 4 CHAPTER I. Linear equations in Linear algebra 0 13.229 872 015-2 01520215900 00 015-2-0103. The solution is(5,3,1) 001 1-30 1-3051「1-30 14 52050 7~01 0~0 10 01100-25700 1-30 100 3051「1002 10-1-0 10-1. The solution is(2, -1, 1) 15. First, replace R4 by R4+(3Rl, then replace r3 by R3+(2)R2, and finally replace r4 by R4+(3)R3 10302 003 303 0327 231 000 0-97-11 0211030 3039 0 2370 00 7-11000 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)RI and replace r4 by r4+(3/2)R2 (One could also scale r2 before adding to R4, but the arithmetic is rather easy keeping R2 unchanged. Finally, replace R4 by R4+R3 100231「10023 0220002200 3100 0 0102200 001 033 100131 00-1 00000 The system is now in triangular form and has a solution. The next section discusses how to continue with this type of system 1.1 Solutio 17. Row reduce the augmented matrix corresponding to the given system of three equations 1-4 1-340-750 470 The system is consistent, and using the argument from Examplc 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 01-1 100 13000 4|000-5 The third equation, 0=-5, shows that the system is inconsistent, so the three planes have no point in common 1h41「1h4 Write c for 6-3h. Ifc=0, that is, if h=2, then the system has no 368|06-3h-4 solution, because 0 cannot equal -4. Otherwise, when h# 2, the system has a solution 1h-3 20 Write c for 4+ 2h. Then the second equation cx2-0 has a solution 24604+2h0 for every value of c. So the system is consistent for all h Write c for h+ 12. Then the second equation cx2=0 has a solution 4h8|0h+120 for every value of c. So the system is consistent for all h 22. The system is consistent if and only if 5+3h=o, that is, if and only 005+3h ifh=-5/3 23. a. True. See the remarks following the box titled Elementary row Operations b. Falsc.a 5x6 matrix has fivc 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 truc 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 paragraph after equation(2) 6 CHAPTER 1. Linear equations in Linear Algebra 7 ghk g 430 g 5 h 259 035k+2g|0 0 k+2g +h Let b denote the number k+ 2g h Then the third equation represented by the augmented matrix above is0=b. This equation is possible if and only if b is zero. So the original system has a solution if and only if k+ 2gt h=o 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 perform any elementary row operations to produce other augmented matrices. Here are three examples. The fact that they are all row equivalent proves that they all have the solution set (2, 1, 0) 100 100 010|201-4 27. Study the augmented matrix for the given system, replacing r2 by r2 +(cRi g」[0d-3cg-cf This shows that shows d-3c must be nonzero, since f and g are arbitrary. Otherwise, for some choices of 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 l/a, which is possible since a is nonzero. Then replace R2 by R2+(CRI abf「1b/af/a bi a f/a g o d-c(b/a)g-c( The quantity d-c(b/)must be nonzero, in order for the system to be consistent when the quantity g-c(fla)is nonzero(which can certainly happen ). The condition that d-c( bla)0 can also be written sad-bc≠0,orad≠b 29. Swap RI and R2; swap rl and r2 30. Multiply r2 by-1/2; multiply R2 by -2 31. Replace R3 by R3+(Rl; replace R3 by R3+(4RI 32. Replace r3 by R3 +(3)R2; replace r3 by R3 +(3)R2 33. The first equation was given. The others are 72=(71+20+40+73)/4,Or412-7i-73 73=(74+72+40+30)/4,Or473-74-2=70 74=(10+T+73+30)/4,or474 T2=40 1.1 Solutions 7 Rearranging, 47i 72 T+472 T T,+4T3 T 70 73+4/4 40 34. Begin by interchanging Rl and R4, then create zeros in the first column 4-10-1301「-10-1440「-10-1440 14-1060 14-1060 040-420 0-14-170 0-14-170 70 4 0-1300-1-415190 Scale rl by-I and r2 by 174, create zeros in the second column, and replace r4 by r4+ r3 101 -401「101-4-401「101-4-40 0-1 14-170004-275004-275 0-1 1519000-41419500012270 Scale r4 by 1712, use R4 to create zeros in column 4, and then scale r3 by 1/4 101-4-40 0 0501「10 50 010-150 002750100275 004 75004012000 030 000 225000 2.50001225 The last step is to replace Rl by R1 +(1R3 100020.0 010027.5 The solution is(20, 27.5, 30, 22. 5) 030.0 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 multiplications or additions(not counting opcrations with zero). This exercise should give students an appreciation for matrix programs such aS MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization For instructors who wish to use technology in the course, the Study guide provides boxed matlaB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and the TI-83+/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 all numerical exercises in the text this feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra The matlab box also explains the basic commands replace, swap and scale. These commands are includedinthetextdatasetsavailablefromthetextwebsitewww.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 470 12 678 235 495 01 10-1 0-5-10-15 341①0-1-2 -o 1 23-0 2 3. Pivot cols 1 and 2. 4567 00000000 6789 7 1357 357 4.3579~048-12~012 2 579 0-8-16-340-8-16-34000-10 0 0-10 0120~0① 2 o Pivot col Is 1,2,and4 3(5)79 000100011000① 579① 60■,0000 000000 7 600 15」001 Corresponding system of equations The basic variables(corresponding to the pivot positions ) are xI and x. The remaining variable x2 is free Solve for the basic variables in terms of the free variable. The gencral solution is 5-3 is free x3=3 Note: Exercise 7 is paired with Exercise 10 1.2 Solutio l40 00 8. 270100-10-401040 Corresponding system of equations The basic variables(corresponding to the pivot positions are xI and x. The remaining variable x3 is free Solve for the basic variables in terms of the free variable. In this particular problem, the basic variables do not depend on the value of the free variable 9 General solution:x,=4 x is free 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.) 01-6 1276①054 7-60 -65」0①-65 sX 4 Corresponding system 4+5 Basic variables: x1, x2; free variable: x3. General solution:x2=5+6x s free 10. 2001-700 2x2 Corresponding system 4+2x Basic variables: xl. xs: free variable: x2. General solution: x, is free ①-4/32/30 11.-912-60~0000~000 684000000000 4 x,+-x Corresponding system 10 CHAPTER 1. Linear Fquations in Linear algebra 4 2 x1 Basic variable:xi: free variables x2, x3. General solution: x, is free x. is free 70651①-7065 12.001-2-3~001-2-3 742700481200000 7 十 Corresponding system 2x 0 5+7x,-6 Basic variables: x1 and x3; free variables: x2, X4. General solution:2 is free 3+2 0-10-21「1-300921000 13. 000194 00194000①9 000000000000 4x Corresponding system C4+ 9xs 5+3 1+4 Basic variables:x1,x2, x4; free variables:x3, xs. General solution:x, is ficc xa= Note: The Study Guide discusses the common mistake x3-0 6-30 -6-302 14 0000100000 ① 00000 0000

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试读 127P Linear Algebra and Its Applications习题解答
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A_Biao_ 是另外一本书的,不是标题上的这本书
2020-12-06
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knwdk 答案很清晰的
2017-10-30
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64位单片鸡 很清晰,但是不对啊,我的书是第四版的。
2016-08-11
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xuelangwin 非常好的一本习题答案
2016-03-26
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yiweiyanghahaha 这是哪本书的??连章节标题都对不上啊
2015-09-26
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maeiei 买了纸质书,答案非常好!
2015-05-12
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duguyuyunxx 答案很清晰的
2014-10-10
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