Solution Manual for Linear Algebra, 6th Edition

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SOLUTIONSMANUALLINEARALGEBRAANDITSAPPLICATIONSSIXTHEDITIONDavid C. LayUniversity of Maryland–College ParkSteven R. LayLee UniversityJudi J. McDonaldWashington State University

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ContentsIntroductionvChapter 1Linear Equations in Linear Algebra1-11.1Systems of Linear Equations1-11.2Row Reduction and Echelon Forms1-81.3Vector Equations1-161.4The Matrix EquationAx=b1-251.5Solution Sets of Linear Systems1-331.6Applications of Linear Systems1-421.7Linear Independence1-511.8Introduction to Linear Transformations1-581.9The Matrix of a Linear Transformation1-651.10Linear Models in Business, Science, and Engineering1-71Supplementary Exercises1-80Chapter 2Matrix Algebra2-12.1Matrix Operations2-12.2The Inverse of a Matrix2-72.3Characterization of Invertible Matrices2-152.4Partitioned Matrices2-232.5Matrix Factorizations2-322.6The Leontief Input-Output Model2-472.7Applications to Computer Graphics2-512.8Subspaces ofn2-582.9Dimension and Rank2-66Supplementary Exercises2-72Chapter 3Determinants3-13.1Introduction to Determinants3-13.2Properties of Determinants3-83.3Cramer’s Rule, Volume, and Linear Transformations3-14Supplementary Exercises3-22Chapter 4Vector Spaces4-14.1Vector Spaces and Subspaces4-14.2Null Spaces, Column Spaces, Row Spaces, and Linear Transformations 4-74.3Linearly Independent Sets; Bases4-154.4Coordinate Systems4-234.5The Dimension of a Vector Space4-304.6Change of Basis4-364.7Digital Signal Processing4-404.8Applications to Difference Equations4-43Supplementary Exercises4-52iii

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Chapter 5Eigenvalues and Eigenvectors5-15.1Eigenvalues and Eigenvectors5-15.2The Characteristic Equation5-105.3Diagonalization5-155.4Eigenvalues and Linear Transformations5-295.5Complex Eigenvalues5-355.6Discrete Dynamical Systems5-435.7Applications to Differential Equations5-495.8Iterative Estimates for Eigenvalues5-595.9Applications to Markov Chains5-67Supplementary Exercises5-75Chapter 6Orthogonality and Least Squares6-16.1Inner Product, Length, and Orthogonality6-16.2Orthogonal Sets6-56.3Orthogonal Projections6-106.4The Gram-Schmidt Process6-186.5Least-Squares Problems6-246.6Machine Learning and Linear Models6-296.7Inner Product Spaces6-346.8Applications of Inner Product Spaces6-38Supplementary Exercises6-43Chapter 7Symmetric Matrices and Quadratic Forms7-17.1Diagonalization of Symmetric Matrices7-17.2Quadratic Forms7-147.3Constrained Optimization7-227.4The Singular Value Decomposition7-277.5Applications to Image Processing and Statistics7-37Supplementary Exercises7-40Chapter 8The Geometry of Vector Spaces8-18.1Affine Combinations8-18.2Affine Independence8-58.3Convex Combinations8-108.4Hyperplanes8-158.5Polytopes8-198.6Curves and Surfaces8-22Supplementary Exercises8-27Chapter 9Optimization9-19.1Matrix Games9-19.2Linear Programming—Geometric Method9-79.3Linear Programming—Simplex Method9-119.4Duality9-15Supplementary Exercises9-22iv

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IntroductionI fell in love with linear algebra when I was an undergraduate student and it has remained acentral part of my life since that time. It is an interesting and beautiful subject, with a broadrange of applications. In recent years, I consistently hear from industry partners about howmuch the high-tech industry appreciates individuals having a strong foundation in bothtechnical and theoretical aspects of linear algebra. I hope you will enjoy teaching this courseas much as I do. You are also welcome to email me at LLinearAlgebra@gmail.com any timeyou have comments and suggestions, or just want to talk about linear algebra.There are many ways in which modern technology can support (or hinder) yourstudent’s learning. The interactive figures from the electronic textbook can be used inclassroom demonstrations to bring linear algebraic concepts alive and demonstrate numerousexamples with the push of a button. Take time to explore with the interactive figures andshow your students how to use technology to find key definitions and theorems quickly in theelectronic textbook.If your course uses MyLab for homework, there are several things to be aware of.First, for some exercises, your students will enter only a final answer. To get to that answer,they may have half a page or more of calculations. Encourage them to keep a notebook withthe exercise statement, worked solutions, and summary notes about what they learned whilesolving an exercise. As an instructor, you can choose many settings in the program. You canset how many tries students are allowed to solve each question. Most exercises let the studenthave three tries before MyLab either records an incorrect answer or offers the student asimilar question. In my experience, persistence pays off – if students are allowed to continueto work similar exercises, mastery of the skill will result. I have also found that the “View anExample” and “Help Me Solve It” tab help get students going again when they are stuck.At the end of each chapter, we have highlighted some of the projects that areavailable online, but moved away from updating the toolbox and the computer manuals. Ifind that when I am trying to code almost anything, I go to the help features for the programor open a search engine and enter some key words. There are still tips in theStudentStudyGuideabout appropriate MATLAB code for various parts of the course.Technology also provides students with easy access to a wealth of videos on linearalgebra and solutions for some of the exercises. Please refrain from posting portions of this’s Solution Manualonline, as by doing so you are giving other instructors’ studentssolutions to the exercises. Some of the open-access online videos are amazing. Otherscontain errors or introduce the material in a different order from how it is covered in this text,leading to confusion. I try to talk to my students about using technology to learn effectivelywithout it becoming a crutch that leaves them with perfect homework and failed exams.The’s Solution Manualcontains detailed solutions for all the exercises,as well as advice on the exercises themselves. I am interested to hear from you atLLinearAlgebra@gmail.com as to what types of material you would like to use in yourcourse, additional topics you would like to see covered, any typos you find, or just to talkabout my favorite subject – linear algebra.—Judi J. McDonaldv

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1-11.1 - Systems Of Linear EquationsNotes:The key exercises are 7 (or 11 or 12), 23–26, and 35. For brevity, the symbols R1, R2,…, stand forrow 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section. InExercises 15–18, students are asked to check their answers to Exercises 11–14; checking that solutions arecorrect, or at least reasonable, is an important skill in the high-tech industry.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).3. The point of intersection satisfies the system of two linear equations:

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1-2Chapter 1Linear Equations in Linear Algebra12125722xxxx+=−= −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 equation1230001xxx++=, or simply, 0 = 1. A system containing this condition has no solution. Furtherrow operations are unnecessary once an equation such as 0 = 1 is evident.The solution set is empty.8. The standard row operations begin by multiplying R3 by ½. Next, replace R2 by R2+ (–7)R3 and R1by R1+ (–2)R3. Finally, replace R1 by R1+ (–1)R2.

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1.1 - Systems Of Linear Equations1-311201120110210050170~0170~0107~01070022001100110011−−−−−The solution set contains one solution:()5, 7,1−−.9. The system has already been reduced to triangular form. Notice that in the last row, the only nonzeroentry is in the last column. The corresponding system of equations does not have a solution. Thesolution set is the empty set.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)R2.120301200010000010400100001000~~001000010000100000100001000010−−−The solution set contains one solution: (0, 0, 0, 0).11.First, swap R1 and R2. Then replace R3 by R3 + (–3)R1. To reach triangular form, replace R3 byR3 + (2)R2. Next, divide the last row by 4.014413321332133213321332~0144~0144~0144~0144375637560241200440011−−−−−−−−−−−−To continue, subtract (4)R3 from R2 and subtract (3)R3 from R1. Finally, replace R1 by R1+(–3)R2.1305100190108010800110011−−−The solution set contains one solution: (19, –8, 1).12. Replace R2 by R2 + (–3)R1 and replace R3 by R3 + (4)R1. To avoid fractions, it is easier to replaceR3 by (–1/6)R3, then interchange R2 and R3.13441344134413443778~0254~0254~0132462406181201320254−−−−−−−−−−−−−−−−−−Next, subtract 2R2 from R3 to complete the forward phase. Replacing R1 by R1+(–4)R3 and R2 byR2+(3)R3 creates zeros above the 1 in the third column. Finally, replace R1 by R1+(3)R213441304100201320102~0102001000100010−−−−−The solution set contains one solution: (2, 2, 0).

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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. The solution found in Exercise 11 is (19, –8, 1). Substituting into the equations23123123443323756xxxxxxxx+= −++= −++=results in( 8)4(1)4193( 8)3(1)23(19)7( 8)5(1)6−+= −+−+= −+−+=, which establishes that the solution found to Exercise 11 isindeed correct.16. The solution found in Exercise 12 is (2, 2, 0). Substituting into the equations12312312334437784624xxxxxxxxx−+= −−+= −−++=results in23(2)4(0)43(2)7(2)7(0)84(2)6(2)2(0)4−+= −−+= −−++=, which establishes that the answer found in Exercise 12 is indeedcorrect.

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1.1 - Systems Of Linear Equations1-517. The solution found in Exercise 13 is (5, 3, –1). Substituting into the equations131232338229752xxxxxxx−=++=+= −results in53( 1)82(5)2(3)9( 1)7(3)5( 1)2−−=++−=+−= −, which establishes that the answer found in Exercise 13 is indeedcorrect.18. The solution found in Exercise 14 is (2, (–1, 1). Substituting into the equations121232335520xxxxxxx−=−++=+=results in23( 1)5(2)( 1)5(1)2( 1)(1)0−−=−+−+=−+=, which establishes that the answer found in Exercise 14 is indeedcorrect.19. First, replace R4 by R4 + (–3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 byR4 + (3)R3.10302103020103301033~023210232130075009711−−−−−−−10302103020103301033~~0034700347009711000510−−−−−−−The resulting triangular system indicates that a solution exists. In fact, using the argument fromExample 2, one can see that the solution is unique.20. 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.21. Row reduce the augmented matrix corresponding to the given system of three equations:

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1-6Chapter 1Linear Equations in Linear Algebra141141141213 ~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.22. 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.23.1414~3680634hhh−−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.24.1313~.2460420hhh−−−+Writecfor 4 + 2h. Then the second equationcx2= 0 has asolution for every value of c. So the system is consistent for allh.25.132132~.480120hh−−−+Writecforh+ 12. Then the second equationcx2= 0 has asolution for every value ofc. So the system is consistent for allh.26.2323~.6950053hhh−−−+The system is consistent if and only if 5 + 3h= 0, that is, if andonly ifh= –5/3.27. True. See the remarks following the box titled “Elementary Row Operations”.28.True. See the box preceding the subsection titled “Existence and Uniqueness Questions”.29. False. A 5 × 6 matrix has five rows.30. False. The definition ofrow equivalentrequires that there exist a sequence of row operations thattransforms one matrix into the other.31. 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.32. False. By definition, an inconsistent system hasnosolution.33. True. See the box before Example 2.34. True. This definition ofequivalent systemsis in the second paragraph after equation (2) in Section1.1 the text.

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1.1 - Systems Of Linear Equations1-735.147147147035~035~03525903520002ggghhhkkgkgh−−−−−−−−−+++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= 0.36. 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 given by(–2, 1, 0).1002100210020101 ~2103 ~2103001000102014−−−−−−37. 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 somechoices offandgthe second row would correspond to an equation of the form 0 =b, wherebisnonzero. Thusd≠3c.38. 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.39. Swap R1 and R2; swap R1 and R2.40. Multiply R2 by –1/2; multiply R2 by –2.41. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1.42. Replace R3 by R3 + (3)R2; replace R3 by R3 + (–3)R2.43. The first equation was given. The others are:213213(2040)/4,or460TTTTTT=+++−−=342342(4030)/4,or470TTTTTT=+++−−=413413(1030)/4,or440TTTTTT=+++−−=Rearranging,

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1-8Chapter 1Linear Equations in Linear Algebra124123234134430460470440TTTTTTTTTTTT−−=−+−=−+−=−−+=44. 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 44 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 not alreadysolved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connectionwith 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 some calculators appear inseparate appendices at the end of theStudy Guide.1.2 - Row Reduction and Echelon FormsNotes:The key exercises are 1–24 and 35–40. (Students should work at least four or five from Exercises7–14, in preparation for Section 1.5.)

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Solution Manual for Linear Algebra, 6th Edition - Page 15 preview image

1.2 - Row Reduction and Echelon Forms1-91. Reduced echelon form:aandc. Echelon form:bandd. Not echelon: none.2. Reduced echelon form:aandb. Echelon form:d. 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= −−=8.1407140714071005~~~27011010301030103−−−Corresponding system of equations:1253xx=−=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.

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Solution Manual for Linear Algebra, 6th Edition - Page 16 preview image

1-10Chapter 1Linear Equations in Linear AlgebraGeneral solution:12353is freexxx= −=Note:A common error in Exercise 8 is to assume thatx3is zero. To avoid this, identify the basic variablesfirst. Any remaining variables arefree. (This type of computation will arise in Chapter 5.)9.016512741056~~127401650165−−−−−−−−Corresponding system:13235665xxxx−=−=Basic variables:x1,x2; free variable:x3. General solution:132336556is 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−+===Basic variable:x1; free variablesx2,x3. General solution:123234233is freeis freexxxxx=−12.17065170651706500123 ~00123 ~001231742700481200000−−−−−−−−−−−−

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