Linear Algebra with Applications, Solutions Manual PDF
Preview Linear Algebra with Applications, Solutions Manual
PREFACE This solutions manual is designed to accompany the eighth edition of Linear Algebra with Applications by Steven J. Leon. The answers in this manual supple- mentthosegivenintheanswerkeyofthetextbook.Inadditionthismanualcontains the complete solutions to allof the nonroutine exercises in the book. At the end of each chapter of the textbook there are two chapter tests (A and B) and a section of computer exercises to be solved using MATLAB. The questionsineachChapterTestAaretobeansweredaseithertrueorfalse.Although the true-false answers are given in the Answer Section of the textbook, students are required to explain or prove their answers. This manual includes explanations, proofs, and counterexamples for all Chapter Test A questions. The chapter tests labeled B contain problems similar to the exercises in the chapter. The answers to these problems are not given in the Answers to Selected Exercises Section of the textbook, however, they are provided in this manual.Complete solutions are given for all of the nonroutine Chapter Test B exercises. IntheMATLABexercises mostofthecomputationsarestraightforward.Con- sequentlythey havenotbeen includedinthissolutionsmanual.Ontheother hand, the text also includes questions related to the computations. The purpose of the questions isto emphasize the significance of the computations.The solutions man- ual does provide the answers to most of these questions. There are some questions for which it is not possible to provide a single answer. For example,some exercises involve randomly generated matrices. In these cases the answers may depend on the particular random matrices that were generated. Steven J. Leon [email protected] Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Contents 1 Matrices and Systems of Equations 1 1 Systems of Linear Equations 1 2 Row Echelon Form 3 3 Matrix Arithmetic 4 4 Matrix Algebra 7 5 Elementary Matrices 13 6 Partitioned Matrices 19 MATLABExercises 23 Chapter Test A 25 Chapter Test B 28 2 Determinants 31 1 The Determinant of a Matrix 31 2 Properties of Determinants 34 3 Additional Topicsand Applications 38 MATLABExercises 40 Chapter Test A 40 Chapter Test B 42 3 Vector Spaces 44 1 Definition and Examples 44 2 Subspaces 49 3 Linear Independence 53 4 Basis and Dimension 57 5 Change of Basis 60 6 Row Space and Column Space 60 MATLABExercises 69 Chapter Test A 70 Chapter Test B 72 iii Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 4 Linear Transformations 76 1 Definition and Examples 76 2 Matrix Representations of Linear Transformations 80 3 Similarity 82 MATLABExercise 84 Chapter Test A 84 Chapter Test B 86 5 Orthogonality 88 1 The Scalar product in Rn 88 2 Orthogonal Subspaces 91 3 Least Squares Problems 94 4 Inner Product Spaces 98 5 Orthonormal Sets 104 6 The Gram-Schmidt Process 113 7 Orthogonal Polynomials 115 MATLABExercises 119 Chapter Test A 120 Chapter Test B 122 6 Eigenvalues 126 1 Eigenvalues and Eigenvectors 126 2 Systems of Linear Differential Equations 132 3 Diagonalization 133 4 HermitianMatrices 142 5 Singular Value Decomposition 150 6 Quadratic Forms 153 7 PositiveDefinite Matrices 156 8 Nonnegative Matrices 159 MATLABExercises 161 Chapter Test A 165 Chapter Test B 167 7 Numerical Linear Algebra 171 1 Floating-Point Numbers 171 2 Gaussian Elimination 171 3 Pivoting Strategies 173 4 Matrix Norms and Condition Numbers 174 5 Orthogonal Transformations 186 6 The Eigenvalue Problem 188 7 Least Squares Problems 192 MATLABExercises 195 Chapter Test A 197 Chapter Test B 198 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 1 Matrices and Systems of Equations 1 SYSTEMS OF LINEAR EQUATIONS 1 1 1 1 1 0 2 1 2 1 − 2. (d) 0 0 4 1 2 − 00 00 00 10 −23 5. (a) 3x1+2x2 =8 x +5x =7 1 2 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 2 Chapter 1 • MatricesandSystemsof Equations (b) 5x 2x + x =3 1 2 3 − 2x +3x 4x =0 1 2 3 − (c) 2x + x +4x = 1 1 2 3 − 4x 2x +3x = 4 1 2 3 − 5x +2x +6x = 1 1 2 2 − (d) 4x 3x + x +2x =4 1 2 3 4 − 3x + x 5x +6x =5 1 2 3 4 − x + x +2x +4x =8 1 2 3 4 5x + x +3x 2x =7 1 2 3 4 − 9. Given the system m x +x = b 1 1 2 1 − m x +x = b 2 1 2 2 − one can eliminate the variable x by subtracting the first row from the 2 second. One then obtains the equivalent system m x +x = b 1 1 2 1 − (m m )x = b b 1 2 1 2 1 − − (a) If m =m , then one can solve the second equation for x 1 2 1 6 b b 2 1 x = − 1 m m 1 2 − One can then plug this value of x into the first equation and solve for 1 x . Thus, if m =m , there willbe a unique ordered pair (x ,x ) that 2 1 2 1 2 6 satisfies the two equations. (b) If m =m , then the x term drops out in the second equation 1 2 1 0=b b 2 1 − This is possible if and only if b =b . 1 2 (c) If m = m , then the two equations represent lines in the plane with 1 2 6 different slopes. Two nonparallel lines intersect in a point. That point willbe the uniquesolutiontothe system. Ifm =m andb =b ,then 1 2 1 2 both equationsrepresent the samelineandconsequently every pointon that line willsatisfy both equations. If m = m and b =b , then the 1 2 1 2 6 equations represent parallel lines. Since parallel lines do not intersect, there is no point on both lines and hence no solution to the system. 10. The system must be consistent since (0,0) is a solution. 11. A linear equation in 3 unknowns represents a plane in three space. The solution set to a 3 3 linear system would be the set of all points that lie × on all three planes. If the planes are parallel or one plane is parallel to the lineofintersection oftheother two,thenthe solutionset willbe empty.The three equations could represent the same plane or the three planes could all intersect in a line. In either case the solution set will contain infinitely many points. If the three planes intersect in a point then the solution set willcontain only that point. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Section2 • Row EchelonForm 3 2 ROW ECHELON FORM 2. (b) The system is consistent with a unique solution (4, 1). − 4. (b) x and x are lead variables and x is a free variable. 1 3 2 (d) x and x are lead variables and x and x are free variables. 1 3 2 4 (f) x and x are lead variables and x is a free variable. 2 3 1 5. (l) The solution is (0, 1.5, 3.5). − − 6. (c) The solution set consists of allordered triples of the form (0, α,α). − 7. A homogeneous linear equation in 3 unknowns corresponds to a plane that passes through the origin in 3-space. Two such equations would correspond to two planes through the origin. If one equation is a multiple of the other, then both represent the same plane through the origin and every point on thatplanewillbeasolutiontothesystem.Ifoneequationisnotamultipleof theother,thenwehavetwodistinctplanesthatintersectinalinethroughthe origin.Every pointonthe lineofintersection willbe asolutiontothe linear system. So in either case the system must have infinitely many solutions. Inthecaseofanonhomogeneous2 3linearsystem,theequationscor- × respond to planes that do not both pass through the origin.If one equation is amultipleof the other, then both represent the same planeand there are infinitelymanysolutions.If the equationsrepresent planesthat are parallel, then theydonotintersect andhence the system willnothaveanysolutions. If the equations represent distinct planes that are not parallel, then they must intersect in a line and hence there will be infinitely many solutions. So the only possibilities for a nonhomogeneous 2 3 linear system are 0 or × infinitely many solutions. 9. (a) Since the system is homogeneous it must be consistent. 13. A homogeneous system is always consistent since it has the trivial solution (0,...,0). If the reduced row echelon form ofthe coefficient matrixinvolves free variables, then there will be infinitely many solutions. If there are no free variables, then the trivial solution willbe the only soltion. 14. A nonhomogeneous system could be inconsistent in which case there would be nosolutions.Ifthe system isconsistent andunderdetermined, then there willbe free variablesand thiswouldimplythat we willhaveinfinitelymany solutions. 16. Ateach intersection the number ofvehicles entering must equal the number ofvehicles leavinginorder forthe traffictoflow.This conditionleadsto the followingsystem of equations x +a = x +b 1 1 2 1 x +a = x +b 2 2 3 2 x +a = x +b 3 3 4 3 x +a = x +b 4 4 1 4 If we add all four equations we get x +x +x +x +a +a +a +a =x +x +x +x +b +b +b +b 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 4 Chapter 1 • MatricesandSystemsof Equations and hence a +a +a +a =b +b +b +b 1 2 3 4 1 2 3 4 17. If (c ,c ) is a solution, then 1 2 a c +a c = 0 11 1 12 2 a c +a c = 0 21 1 22 2 Multiplyingboth equations through by α, one obtains a (αc )+a (αc ) = α 0=0 11 1 12 2 · a (αc )+a (αc ) = α 0=0 21 1 22 2 · Thus (αc ,αc ) is also a solution. 1 2 18. (a)Ifx =0thenx ,x ,andx willallbe0.Thus ifnoglucoseisproduced 4 1 2 3 then there is no reaction. (0,0,0,0)is the trivial solution in the sense that ifthere are no molecules of carbon dioxide and water, then there willbe no reaction. (b) If we choose another value of x , say x = 2, then we end up with 4 4 solution x =12, x =12, x =12,x =2. Note the ratios are still 6:6:6:1. 1 2 3 4 3 MATRIX ARITHMETIC 8 15 11 − 1. (e) 0 4 3 − − 1 6 6 −5 −1−0 15 (g) 5 1 4 − 8 9 6 − 2. (d) 36 10 56 10 3 16 15 20 5. (a) 5A= 5 5 10 35 6 8 9 12 15 20 2A+3A= 2 2 + 3 3 = 5 5 4 14 6 21 10 35 18 24 (b) 6A= 6 6 12 42 6 8 18 24 3(2A)=3 2 2 = 6 6 4 14 12 42 (c) AT = 31 2 4 1 7 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Section3 • Matrix Arithmetic 5 T 3 4 3 1 2 (AT)T = = 1 1 =A 4 1 7 2 7 6. (a) A+B =5 4 6=B+A 0 5 1 (b) 3(A+B)=3 5 4 6 = 15 12 18 0 5 1 0 15 3 3A+3B = 12 3 18+ 3 9 0 6 9 15 6 6 12 − − 15 12 18 = 0 15 3 5 4 6 T 5 0 (c) (A+B)T = = 4 5 0 5 1 6 1 4 2 1 −2 5 0 AT +BT = 1 3 + 3 2 = 4 5 6 5 0 4 6 1 − 5 14 15 42 7. (a) 3(AB) =3 15 42 = 45 126 0 16 0 48 6 3 24 15 42 (3A)B = 18 9 = 45 126 1 6 6 12 0 48 −2 1 6 12 15 42 A(3B) = 6 3 = 45 126 3 18 2 4 0 48 −5 14T 5 15 0 (b) (AB)T = 15 42 = 14 42 16 0 16 BTAT = 2 1 2 6 −2 = 5 15 0 4 6 1 3 4 14 42 16 05 3 1 3 6 8. (a) (A+B)+C = + = 1 7 2 1 3 8 A+(B+C)=2 4+1 2=3 6 1 3 2 5 3 8 (b) (AB)C = −4 18 3 1 = 24 14 2 13 2 1 20 11 − 2 4 4 1 24 14 A(BC) = − − = 1 3 8 4 20 11 (c) A(B+C)= 2 4 1 2 = 10 24 1 3 2 5 7 17 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 6 Chapter 1 • MatricesandSystemsof Equations 4 18 14 6 10 24 AB+AC = − + = 2 13 9 4 7 17 − (d) (A+B)C =0 5 31 = 10 5 1 7 2 1 17 8 14 6 4 1 10 5 AC+BC = + − − = 9 4 8 4 17 8 9. (b)x=(2,1)T isasolutionsince b=2a +a .Therearenoother solutions 1 2 since the echelon form of A is strictly triangular. (c) The solution to Ax=c is x=( 5, 1)T. Therefore c= 5a 1a . −2 −4 −2 1− 4 2 11. The given informationimplies that 1 0 x = 1 and x = 1 1 2 0 1 arebothsolutionstothesystem.Sothesystemisconsistentandsincethereis morethanonesolutiontherowechelonformofAmustinvolveafreevariable. A consistent system with a free variable has infinitely many solutions. 12. The system is consistent since x=(1,1,1,1)T is asolution.The system can haveat most 3 lead variablessince A onlyhas 3 rows. Therefore there must be at least one free variable. A consistent system with a free variable has infinitely many solutions. 13. (a) It followsfromthe reduced row echelon formthat the free variablesare x , x , x . If we set x =a, x =b, x =c, then 2 4 5 2 4 5 x = 2 2a 3b c 1 − − − − x = 5 2b 4c 3 − − and hence the solution consists of all vectors of the form x=( 2 2a 3b c, a, 5 2b 4c, b, c)T − − − − − − (b) If we set the free variables equal to 0, then x = ( 2,0,5,0,0)T is a 0 − solution to Ax=b and hence b=Ax = 2a +5a =(8, 7, 1,7)T 0 1 3 − − − 14. AT is an n m matrix. Since AT has m columns and A has m rows, the × multiplicationATA is possible. The multiplicationAAT is possible since A has n columns and AT has n rows. 15. If A is skew-symmetric then AT = A. Since the (j,j) entry of AT is a jj − and the (j,j) entry of A is a , it follows that is a = a for each j jj jj jj − − − and hence the diagonalentries of A must all be 0. 16. The search vector is x = (1,0,1,0,1,0)T. The search result is given by the vector y=ATx=(1,2,2,1,1,2,1)T The ith entry of y is equalto the number of search words in the title of the ith book. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Section4 • Matrix Algebra 7 17. If α=a /a , then 21 11 1 0 a a a a a a 11 12 = 11 12 = 11 12 α 1 0 b αa αa +b a αa +b 11 12 21 12 The productwillequal A provided αa +b=a 12 22 Thus we must choose a a 21 12 b=a αa =a 22 12 22 − − a 11 4 MATRIX ALGEBRA 1. (a) (A+B)2 =(A+B)(A+B) =(A+B)A+(A+B)B =A2+BA+AB+B2 In the case of real numbers ab + ba = 2ab, however, with matrices AB+BA is generally not equal to 2AB. (b) (A+B)(A B) = (A+B)(A B) − − = (A+B)A (A+B)B − = A2+BA AB B2 − − Inthecaseofrealnumbersab ba=0,however,withmatricesAB BA − − is generally not equal to O. 2. If we replace a by A and b by the identity matrix, I, then both rules will work, since (A+I)2 =A2+IA+AI +B2 =A2+AI +AI +B2 =A2+2AI +B2 and (A+I)(A I)=A2+IA AI I2 =A2+A A I2 =A2 I2 − − − − − − 3. Therearemanypossiblechoices forAandB.Forexample,onecouldchoose 0 1 1 1 A= and B = 0 0 0 0 More generally if a b db eb A= B = ca cb da ea − − then AB =O for any choice ofthe scalars a, b, c, d, e. 4. Toconstruct nonzeromatricesA,B,C withthedesired properties, firstfind nonzero matrices C and D such that DC = O (see Exercise 3). Next, for any nonzero matrix A, set B =A+D. It followsthat BC =(A+D)C =AC+DC =AC+O =AC Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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