Solution Manual for Linear Algebra and Its Applications, 5th Edition

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’SSOLUTIONSMANUALJUDIJ.MCDONALDWashington State UniversityLINEARALGEBRAANDITSAPPLICATIONSFIFTHEDITIONDavid C. LayUniversity of MarylandSteven R. LayLee UniversityJudi J. McDonaldWashington State University

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1-11.1SOLUTIONSNotes:The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, standfor row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section.1.121257275xxxx 157275Replace R2 by R2 + (2)R1 and obtain:1225739xxx157039Scale R2 by 1/3:122573xxx157013Replace R1 by R1 + (–5)R2:1283xx 108013The solution is (x1,x2) = (–8, 3), or simply (–8, 3).2.12122445711xxxx 2445711Scale R1 by 1/2 and obtain:1212225711xxxx 1225711Replace R2 by R2 + (–5)R1:12222321xxx 1220321Scale R2 by –1/3:122227xxx  122017Replace R1 by R1 + (–2)R2:12127xx 1012017The solution is (x1,x2) = (12, –7), or simply (12, –7).

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1-2CHAPTER 1•Linear Equations in Linear Algebra3. The point of intersection satisfies the system of two linear equations:12125722xxxx 157122Replace R2 by R2 + (–1)R1 and obtain:1225779xxx 157079Scale R2 by –1/7:122579/7xxx157019/7Replace R1 by R1 + (–5)R2:124/79/7xx104/7019/7The point of intersection is (x1,x2) = (4/7, 9/7).4. The point of intersection satisfies the system of two linear equations:121251375xxxx151375Replace R2 by R2 + (–3)R1 and obtain:1225182xxx151082Scale R2 by 1/8:122511/4xxx151011/4Replace R1 by R1 + (5)R2:129/41/4xx109/4011/4The point of intersection is (x1,x2) = (9/4, 1/4).5. The system is already in “triangular” form. The fourth equation isx4= –5, and the other equations donot contain the variablex4. The next two steps should be to use the variablex3in the third equation toeliminate that variable from the first two equations. In matrix notation, that means to replace R2 byits 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, whichproduces164010270400123000515. 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 thirdcolumn. But in this case, the third row of the augmented matrix corresponds to the equation 0x1+ 0x2+ 0x3= 1, or simply, 0 = 1. A system containing this condition has no solution. Further rowoperations are unnecessary once an equation such as 0 = 1 is evident.The solution set is empty.

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1.1•Solutions1-38. The standard row operations are:14901490140010000170~0170~0100~01000020001000100010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 andthen replacing R3 by R3 + (3)R4:110041100411004013070130701307~~001310013100105000240001200012Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2:11004100040100801008~~00105001050001200012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 thefinal step, replace R1 by R1 + (2)R2.120321200710003010470100501005~~001060010600106000130001300013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)R2.01451352135213521352~0145~0145~014537763776028120002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)R2.1344134413443778~0254~02544617061590003The system is inconsistent, because the last row would require that 0 = 3 if there were a solution.The solution set is empty.

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1-4CHAPTER 1•Linear Equations in Linear Algebra13. Replace R2 by R2 + (–2)R1. Then interchange R2 and R3. Next replace R3 by R3 + (–2)R2. Thendivide R3 by 5. Finally, replace R1 by R1 + (–2)R3.10381038103810382297~02159~0152~0152015201520215900551038~015200111005~01030011. The solution is (5, 3, –1).14. Replace R2 by R2 + R1. Then interchange R2 and R3. Next replace R3 by R3 + 2R2. Then divideR3 by 7.Next replace R2 by R2 + (–1)R3. Finally, replace R1 by R1 + 3R2.13051305130513051152~0257~0110~011001100110025700771305~0110001113051002~0101 ~010100110011. 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 byR4 + (3)R3.10302103020103301033~0232102321300750097111030201033~003470097111030201033~00347000510.The resulting triangular system indicates that a solution exists. In fact, using the argument fromExample 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 +R3.10023100230220002200~0013100131232150323110023100230220002200~~00131001310013100000The system is now in triangular form and has a solution. The next section discusses how to continuewith this type of system.

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1.1•Solutions1-517.Row reduce the augmented matrix corresponding to the given system of three equations:141141141213 ~075~075134075000The system is consistent, and using the argument from Example 2, there is only one solution. So thethree lines have only one point in common.18.Row reduce the augmented matrix corresponding to the given system of three equations:1214121412140111 ~0111 ~0111130001140005The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point incommon.19.1414~3680634hhhWritecfor 6 – 3h. Ifc= 0, that is, ifh= 2, then the system has nosolution, because 0 cannot equal –4. Otherwise, whenh2, the system has a solution.20.1313~.2460420hhhWritecfor 4 + 2h. Then the second equationcx2= 0 has asolution for every value of c. So the system is consistent for allh.21.132132~.480120hhWritecforh+ 12. Then the second equationcx2= 0 has asolution for every value ofc. So the system is consistent for allh.22.2323~.6950053hhhThe system is consistent if and only if 5 + 3h= 0, that is, if andonly ifh= –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 applies to a single solution. The solutionsetconsists of all possiblesolutions. Only in special cases does the solution set consist of exactly one solution. Mark astatement True only if the statement isalwaystrue.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 ofrow equivalentrequires that there exist a sequence of row operations thattransforms one matrix into the other.c. False. By definition, an inconsistent system hasnosolution.d. True. This definition ofequivalent systemsis in the second paragraph after equation (2).

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1-6CHAPTER 1•Linear Equations in Linear Algebra25.147147147035~035~03525903520002ggghhhkkgkghLetbdenote the numberk+ 2g+h. Then the third equation represented by the augmented matrixabove is 0 =b. This equation is possible if and only ifbis zero. So the original system has a solutionif and only ifk+ 2g+h= 0.26.A basic principle of this section is that row operations do not affect the solution set of a linearsystem. Begin with a simple augmented matrix for which the solution is obviously–2, 1, 0, andthen perform any elementary row operations to produce other augmented matrices. Here are threeexamples. The fact that they are all row equivalent proves that they all have the solution set–2, 1, 0 .1002100210020101 ~2103 ~210300100010201427.Study the augmented matrix for the given system, replacing R2 by R2 + (–c)R1:1313~03ffcdgdcgcf. This shows that showsd– 3cmust be nonzero, sincefandgare arbitrary. Otherwise, for some choices offandgthe second row would correspond to an equationof the form 0 =b, wherebis nonzero. Thusd3c.28.Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possiblesinceais nonzero. Then replace R2 by R2 + (–c)R1.1//1//~~0(/)(/)abfb afab afacdgcdgdc b agc faThe quantityd–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 thatd–c(b/a)0 can also bewritten asad–bc0, orad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:213213(2040)/4,or460TTTTTT342342(4030)/4,or470TTTTTT413413(1030)/4,or440TTTTTT

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1.1•Solutions1-7Rearranging,124123234134430460470440TTTTTTTTTTTT34.Begin by interchanging R1 and R4, then create zeros in the first column:410130101440101440141060141060040420~~01417001417001417010144041013001415190Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3:101440101440101440010150101501015~~~014170004275004275014151900041419500012270Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4:10144010105010105001015010027.5010027.5~~~0042750040120001030000122.5000122.5000122.5The last step is to replace R1 by R1 + (–1)R3:100020.0010027.5~.001030.0000122.5The solution is (20, 27.5, 30, 22.5).Notes:TheStudy Guideincludes 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 operationswith 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 notalready solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem inconnection with an LU factorization.For instructors who wish to use technology in the course, theStudy Guideprovides boxed MATLABnotes at the ends of many sections. Parallel notes for Maple, Mathematica, and ssome calculators appearin separate appendices at the end of theStudy Guide. The MATLAB box for Section 1.1 describes how toaccess the data that is available for all numerical exercises in the text. This feature has the ability to savestudents time if they regularly have their matrix program at hand when studying linear algebra. TheMATLAB box also explains the basic commandsreplace,swap, andscale. These commands areincluded in the text data sets, available from the text web site, www.pearsonhighered.com/lay.

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1-8CHAPTER 1•Linear Equations in Linear Algebra1.2SOLUTIONSNotes:The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises7–14, in preparation for Section 1.5.)1. Reduced echelon form:aandb. Echelon form:d. Not echelon:c.2. Reduced echelon form:a. Echelon form:bandd. Not echelon:c.3.1234123412344567~0369~0123678905101505101512341012~0123~012300000000. Pivot cols 1 and 2.1234456767894.13571357135713573579~04812~0123~0123579108163408163400010135713501010~0123 ~0120~0120000100010001.Pivot cols1, 2, and 41357357957915.**0,,00000      6.**00,00 , 00000000          7.1347134713471305~~~39760051500130013Corresponding system of equations:123353xxxThe basic variables (corresponding to the pivot positions) arex1andx3. The remaining variablex2isfree. Solve for the basic variables in terms of the free variable. The general solution is122353is free3xxxx Note:Exercise 7 is paired with Exercise 10.

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Solution Manual for Linear Algebra and Its Applications, 5th Edition - Page 11 preview image

1.2•Solutions1-98.1407140714071009~~~27010010401040104Corresponding system of equations:1294xxThe basic variables (corresponding to the pivot positions) arex1andx2. The remaining variablex3isfree. Solve for the basic variables in terms of the free variable. In this particular problem, the basicvariables do not depend on the value of the free variable.General solution:12394is freexxx Note:A common error in Exercise 8 is to assume thatx3is zero. To avoid this, identify the basicvariables first. Any remaining variables arefree. (This type of computation will arise in Chapter 5.)9.016512761054~~127601650165Corresponding system:13235465xxxxBasic variables:x1,x2; free variable:x3. General solution:132334556is freexxxxx10.121312131204~~362200170017Corresponding system:123247xxxBasic variables:x1,x3; free variable:x2. General solution:122342is free7xxxx  11.3420342014 / 32 / 3091260~0000~0000684000000000Corresponding system:123420330000xxx

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Solution Manual for Linear Algebra and Its Applications, 5th Edition - Page 12 preview image

1-10CHAPTER 1•Linear Equations in Linear AlgebraBasic variable:x1; free variablesx2,x3. General solution:123234233is freeis freexxxxx12.17065170651706500123~00123~001231742700481200000Corresponding system:124347652300xxxxxBasic variables:x1andx3; free variables:x2,x4. General solution:1242344576is free32is freexxxxxxx 13.130102130092100035010041010041010041~~000194000194000194000000000000000000Corresponding system:15254535419400xxxxxxBasic variables:x1,x2,x4; free variables:x3,x5. General solution:152534555314is free49is freexxxxxxxxNote:TheStudy Guidediscusses the common mistakex3= 0.14.125605107009016302016302~000010000010000000000000

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Solution Manual for Linear Algebra and Its Applications, 5th Edition - Page 13 preview image

1.2•Solutions1-11Corresponding system:13234579632000xxxxxxBasic variables:x1,x2,x5; free variables:x3,x4. General solution:1323434597263is freeis free0xxxxxxxx 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 becausex2is a free variable.17.2323~4670072hhhThe system has a solution only if 7 – 2h= 0, that is, ifh= 7/2.18.132132~570153hhIfh +15 is zero, that is, ifh= –15, then the system has nosolution, because 0 cannot equal 3. Otherwise, when15,h the system has a solution.19.1212~480848hhkhka. Whenh= 2 and8,kthe augmented column is a pivot column, and the system is inconsistent.b. When2,hthe system is consistent and has a unique solution. There are no free variables.c. Whenh= 2 andk= 8, the system is consistent and has many solutions.20.132132~3096hkhka. Whenh= 9 and6,kthe system is inconsistent, because the augmented column is a pivotcolumn.b. When9,hthe system is consistent and has a unique solution. There are no free variables.c. Whenh= 9 andk= 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.

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Solution Manual for Linear Algebra and Its Applications, 5th Edition - Page 14 preview image

1-12CHAPTER 1•Linear Equations in Linear Algebra22.a. False. See the statement preceding Theorem 1. Only thereducedechelon 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 freevariables. If the system is inconsistent, the solution set is empty. See the solution of PracticeProblem 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)row of the coefficient matrix. The reduced echelon form cannot contain a row of the form[000001].24. The system is inconsistent because the pivot in column 5 means that there is a row of the form[00001] in the reduced echelon form. Since the matrix is theaugmentedmatrix for a system,Theorem 2 shows that the system has 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 form123100010and so001axabxbcxcNo matter what the values ofa,b, andc, 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 thecoefficient matrix is a pivot column; otherwise there are infinitely many solutions. ”This statement is true because the free variables correspond tononpivotcolumns of the coefficientmatrix. The columns are all pivot columns if and only if there are no free variables. And there are nofree variables if and only if the solution is unique, by Theorem 2.28. Every column in the augmented matrixexcept the rightmost columnis a pivot column, and therightmost column isnota 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.30. Example:12312342225xxxxxx31. Yes, a system of linear equations with more equations than unknowns can be consistent.

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1.2•Solutions1-13Example (in whichx1=x2= 1):12121220325xxxxxx32. According to the numerical note in Section 1.2, whenn= 30 the reduction to echelon form takesabout 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%.Whenn= 300, the estimates are 2(300)3/3 = 18,000,000 phase for the reduction to echelon formand (300)2= 90,000 flops for the backward phase. The fraction associated with the backward phaseis about (9×104) /(18×106) = .005, or about .5%.33. For a quadratic polynomialp(t) =a0+a1t+a2t2to exactly fit the data (1, 12), (2, 15), and (3, 16), thecoefficientsa0,a1,a2must satisfy the systems of equations given in the text. Row reduce theaugmented matrix:1111211112111121111212415~0133~0133~01331391602840022001111013~010600111007~01060011The polynomial isp(t) = 7 + 6t–t2.34.[M]The system of equations to be solved is:23450123452345012345234501234523450123452345012345230123000000222222.904444414.86666639.68888874.3101010aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa45451010119aaThe unknowns area0,a1, …,a5. Use technology to compute the reduced echelon of the augmentedmatrix:234510000001000000124816322.9024816322.9141664256102414.8008482249609~16362161296777639.600241921248768030.9186451240963276874.3004848040323264062.7008096099209984010110101010101194.5

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Solution Manual for Linear Algebra and Its Applications, 5th Edition - Page 16 preview image

1-14CHAPTER 1•Linear Equations in Linear Algebra10000001000000024816322.9024816322.9008482249609008482249609~~0004857648003.90004857648003.90001922688268808.7000038476806.900048076809024014.5000019204224024.510000001000000024816322.9024816322.9008482249609008482249609~~0004857648003.90004857648003.9000038476806.9000038476806.900000384010000001.00261000000100000002481602.81670100001.71250084822406.50000010001.1948~~~0004857608.6000000100.66150000384026.900000010.0701000001.002604000001.0026Thusp(t) = 1.7125t– 1.1948t2+ .6615t3– .0701t4+ .0026t5, andp(7.5) = 64.6 hundred lb.Notes:In Exercise 34, if the coefficients are retained to higher accuracy than shown here, thenp(7.5) =64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. Theaugmented matrix for such a system is the same as the one used to findp, except that at least column 6 ismissing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will beentirely 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-levelcommands such asgaussandbgauss, discussed in Section 1.4 of theStudy Guide.The commandref(reduced echelon form) is available, but I recommend postponing that command until Chapter 2.TheStudy Guideincludes a “Mathematical Note” about the phrase, “if and only if,” used in Theorem2.1.3SOLUTIONSNotes:The key exercises are 11–14, 17–22, 25, and 26. A discussion of Exercise 25 will help studentsunderstand the notation [a1a2a3], {a1,a2,a3}, and Span{a1,a2,a3}.1.131( 3)4212( 1)1  uv.Using the definitions carefully,131( 2)( 3)1652( 2)212( 2)( 1)224    uv, or, more quickly

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