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Solution Manual for Linear Algebra with Applications, 10th Edition

Solution Manual for Linear Algebra with Applications, 10th Edition is your guide to textbook mastery, offering detailed solutions to every chapter's exercises.

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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 1 preview imageSOLUTIONSMANUALLINEARALGEBRAWITHAPPLICATIONSTENTHEDITIONSteven J. LeonUniversity of Massachusetts DartmouthLisette G. de PillisHarvey Mudd College
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 2 preview image
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 3 preview imageContentsPrefacev1Matrices and Systems of Equations11Systems of Linear Equations12Row Echelon Form23Matrix Arithmetic34Matrix Algebra75Elementary Matrices126Partitioned Matrices18MATLAB Exercises22Chapter Test A24Chapter Test B272Determinants301The Determinant of a Matrix302Properties of Determinants333Additional Topics and Applications36MATLAB Exercises38Chapter Test A38Chapter Test B393Vector Spaces421Definition and Examples422Subspaces463Linear Independence524Basis and Dimension555Change of Basis576Row Space and Column Space57MATLAB Exercises65Chapter Test A66Chapter Test B684Linear Transformations721Definition and Examples722Matrix Representations of Linear Transformations753Similarity78MATLAB Exercise79iii
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 4 preview imageivContentsChapter Test A80Chapter Test B815Orthogonality841The Scalar product inRn842Orthogonal Subspaces863Least Squares Problems894Inner Product Spaces935Orthonormal Sets996The Gram-Schmidt Process1077Orthogonal Polynomials109MATLAB Exercises112Chapter Test A113Chapter Test B1156Eigenvalues1191Eigenvalues and Eigenvectors1192Systems of Linear Differential Equations1243Diagonalization1254Hermitian Matrices1335Singular Value Decomposition1416Quadratic Forms1437Positive Definite Matrices1468Nonnegative Matrices149MATLAB Exercises152Chapter Test A154Chapter Test B1567Numerical Linear Algebra1601Floating-Point Numbers1602Gaussian Elimination1613Pivoting Strategies1624Matrix Norms and Condition Numbers1635Orthogonal Transformations1746The Eigenvalue Problem1757Least Squares Problems1798Iterative Methods182MATLAB Exercises183Chapter Test A185Chapter Test B186
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 5 preview imagePrefaceThis solutions manual is designed to accompany the tenth edition ofLinear Algebra with Applicationsby Steven J. Leon and Lisette de Pillis. The manual contains the complete solutions to all of thenonroutine exercises and Chapter test questions in the first seven chapters the book. Each of thosechapters also includes a set of MATLAB computer exercises. Most of the MATLAB computationsare straightforward. and consequently the computational results are not included in this manual.However, the MATLAB Exercises also include questions related to the computations. The purposeof the questions is to emphasize the significance of the computations. This manual does provide theanswers to most of these questions.v
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 6 preview imageChapter 1Matrices andSystemsof Equations1SYSTEMS OF LINEAR EQUATIONS2.(d)11111021−210041−20001−3000025.(a) 3x1+ 2x2= 8x1+ 5x2= 7(b) 5x1−2x2+x3= 32x1+ 3x2−4x3= 0(c) 2x1+x2+ 4x3=−14x1−2x2+ 3x3=45x1+ 2x2+ 6x2=−11
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 7 preview image2Chapter 1•Matrices and Systems of Equations(d) 4x1−3x2+x3+ 2x4= 43x1+x2−5x3+ 6x4= 5x1+x2+ 2x3+ 4x4= 85x1+x2+ 3x3−2x4= 79.Given the system−m1x1+x2=b1−m2x1+x2=b2one can eliminate the variablex2by subtracting the first row from the second. One thenobtains the equivalent system−m1x1+x2=b1(m1−m2)x1=b2−b1(a) Ifm16=m2, then one can solve the second equation forx1x1=b2−b1m1−m2One can then plug this value ofx1into the first equation and solve forx2. Thus, ifm16=m2, there will be a unique ordered pair (x1, x2) that satisfies the two equations.(b) Ifm1=m2, then thex1term drops out in the second equation0 =b2−b1This is possible if and only ifb1=b2.(c) Ifm16=m2, then the two equations represent lines in the plane with different slopes.Two nonparallel lines intersect in a point. That point will be the unique solution tothe system. Ifm1=m2andb1=b2, then both equations represent the same line andconsequently every point on that line will satisfy both equations. Ifm1=m2andb16=b2,then the equations represent parallel lines. Since parallel lines do not intersect, there isno 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 3×3linear system would be the set of all points that lie on all three planes. If the planes areparallel or one plane is parallel to the line of intersection of the other two, then the solutionset will be empty. The three equations could represent the same plane or the three planescould 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 will contain only that point.2ROW ECHELON FORM2.(b) The system is consistent with a unique solution (4,−1).4.(b)x1andx3are lead variables andx2is a free variable.(d)x1andx3are lead variables andx2andx4are free variables.(f)x2andx3are lead variables andx1is a free variable.5.(l) The solution is (0,−1.5,−3.5).6.(c) The solution set consists of all ordered triples of the form (0,−α, α).7.A homogeneous linear equation in 3 unknowns corresponds to a plane that passes throughthe 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 theorigin and every point on that plane will be a solution to the system. If one equation is nota multiple of the other, then we have two distinct planes that intersect in a line through the
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 8 preview imageSection 3•Matrix Arithmetic3origin. Every point on the line of intersection will be a solution to the linear system. So ineither case the system must have infinitely many solutions.In the case of a nonhomogeneous 2×3 linear system, the equations correspond to planesthat do not both pass through the origin. If one equation is a multiple of the other, then bothrepresent the same plane and there are infinitely many solutions. If the equations representplanes that are parallel, then they do not intersect and hence the system will not have anysolutions. If the equations represent distinct planes that are not parallel, then they mustintersect in a line and hence there will be infinitely many solutions. So the only possibilitiesfor 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 thereduced row echelon form of the coefficient matrix involves free variables, then there will beinfinitely many solutions. If there are no free variables, then the trivial solution will be theonly solution.14.A nonhomogeneous system could be inconsistent in which case there would be no solutions.If the system is consistent and underdetermined, then there will be free variables and thiswould imply that we will have infinitely many solutions.16.At each intersection, the number of vehicles entering must equal the number of vehicles leavingin order for the traffic to flow. This condition leads to the following system of equationsx1+a1=x2+b1x2+a2=x3+b2x3+a3=x4+b3x4+a4=x1+b4If we add all four equations, we getx1+x2+x3+x4+a1+a2+a3+a4=x1+x2+x3+x4+b1+b2+b3+b4and hencea1+a2+a3+a4=b1+b2+b3+b417.If (c1, c2) is a solution, thena11c1+a12c2= 0a21c1+a22c2= 0Multiplying both equations through byα, one obtainsa11(αc1) +a12(αc2) =α·0 = 0a21(αc1) +a22(αc2) =α·0 = 0Thus (αc1, αc2) is also a solution.18.(a) Ifx4= 0, thenx1,x2, andx3will all be 0. Thus if no glucose is produced, then thereis no reaction. (0,0,0,0) is the trivial solution in the sense that if there are no molecules ofcarbon dioxide and water, then there will be no reaction.(b) If we choose another value ofx4, sayx4= 2, then we end up with solutionx1= 12,x2= 12,x3= 12,x4= 2. Note the ratios are still 6:6:6:1.3MATRIX ARITHMETIC1.(e)8−15110−4−3−1−66
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 9 preview image4Chapter 1•Matrices and Systems of Equations(g)5−10155−148−962.(d)361056103165.(a) 5A=15205510352A+ 3A=6822414+91233621=1520551035(b) 6A=18246612423(2A) = 36822414=1824661242(c)AT=312417(AT)T=312417T=341127=A6.(a)A+B=546051=B+A(b) 3(A+B) = 3546051=15121801533A+ 3B=123186915+390−66−12=1512180153(c) (A+B)T=546051T=504561
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 10 preview imageSection 3•Matrix Arithmetic5AT+BT=421365+1−2320−4=5045617.(a) 3(AB) = 35141542016=154245126048(3A)B=63189−6122416=154245126048A(3B) =2163−24612318=154245126048(b) (AB)T=5141542016T=5150144216BTAT=214626−2134=51501442168.(a) (A+B) +C=0517+3121=3638A+ (B+C) =2413+1225=3638(b) (AB)C=−418−2133121=24142011A(BC) =2413−4−184=24142011(c)A(B+C) =24131225=1024717AB+AC=−418−213+14694=1024717(d) (A+B)C=05173121=105178AC+BC=14694+−4−184=1051789.(b)x= (2,1)Tis a solution sinceb= 2a1+a2. There are no other solutions since the echelonform ofAis strictly triangular.(c) The solution toAx=cisx= (−52,−14)T. Thereforec=−52a1−14a2.
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 11 preview image6Chapter 1•Matrices and Systems of Equations11.The given information implies thatx1=110andx2=011are both solutions to the system. So the system is consistent and since there is more than onesolution, the row echelon form ofAmust involve a free variable. A consistent system with afree variable has infinitely many solutions.12.The system is consistent sincex= (1,1,1,1)Tis a solution. The system can have at most 3lead variables sinceAonly has 3 rows. Therefore, there must be at least one free variable. Aconsistent system with a free variable has infinitely many solutions.13.(a) It follows from the reduced row echelon form that the free variables arex2,x4,x5. If wesetx2=a,x4=b,x5=c, thenx1=−2−2a−3b−cx3= 5−2b−4cand hence the solution consists of all vectors of the formx= (−2−2a−3b−c, a,5−2b−4c, b, c)T(b) If we set the free variables equal to 0, thenx0= (−2,0,5,0,0)Tis a solution toAx=band henceb=Ax0=−2a1+ 5a3= (8,−7,−1,7)T14.Ifw3is the weight given to professional activities, then the weights for research and teachingshould bew1= 3w3andw2= 2w3. Note that1.5w2= 3w3=w1,so the weight given to research is 1.5 times the weight given to teaching. Since the weightsmust all add up to 1, we have1 =w1+w2+w3= 3w3+ 2w3+w3= 6w3and hence it follows thatw3=16,w2=13,w1=12. IfCis the matrix in the example problemfrom the Analytic Hierarchy Process Application, then the rating vectorris computed bymultiplyingCtimes the weight vectorw.r=Cw=1215141412121431014121316=43120451203212015.ATis ann×mmatrix. SinceAThasmcolumns andAhasmrows, the multiplicationATAis possible. The multiplicationAATis possible sinceAhasncolumns andAThasnrows.16.IfAis skew-symmetric, thenAT=−A. Since the (j, j) entry ofATisajjand the (j, j) entryof−Ais−ajj, it follows thatajj=−ajjfor eachjand hence the diagonal entries ofAmustall be 0.17.The search vector isx= (1,0,1,0,1,0)T. The search result is given by the vectory=ATx= (1,2,2,1,1,2,1)T
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 12 preview imageSection 4•Matrix Algebra7Theith entry ofyis equal to the number of search words in the title of theith book.18.Ifα=a21/a11, then10α1a11a120b=a11a12αa11αa12+b=a11a12a21αa12+bThe product will equalAprovidedαa12+b=a22Thus we must chooseb=a22−αa12=a22−a21a12a114MATRIX ALGEBRA1.(a) (A+B)2= (A+B)(A+B) = (A+B)A+ (A+B)B=A2+BA+AB+B2For real numbers,ab+ba= 2ab; however, with matricesAB+BAis generally not equalto 2AB.(b)(A+B)(A−B) = (A+B)(A−B)= (A+B)A−(A+B)B=A2+BA−AB−B2For real numbers,ab−ba= 0; however, with matricesAB−BAis generally not equaltoO.2.If we replaceabyAandbby the identity matrix,I, then both rules will work, since(A+I)2=A2+IA+AI+B2=A2+AI+AI+B2=A2+ 2AI+B2and(A+I)(A−I) =A2+IA−AI−I2=A2+A−A−I2=A2−I23.There are many possible choices forAandB. For example, one could chooseA=0100andB=1100More generally ifA=abcacbB=dbeb−da−eathenAB=Ofor any choice of the scalarsa,b,c,d,e.4.To construct nonzero matricesA,B,Cwith the desired properties, first find nonzero matricesCandDsuch thatDC=O(see Exercise 3). Next, for any nonzero matrixA, setB=A+D.It follows thatBC= (A+D)C=AC+DC=AC+O=AC5.A 2×2 symmetric matrix is one of the formA=abbc
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 13 preview image8Chapter 1•Matrices and Systems of EquationsThusA2=a2+b2ab+bcab+bcb2+c2IfA2=O, then its diagonal entries must be 0.a2+b2= 0andb2+c2= 0Thusa=b=c= 0 and henceA=O.6.LetD= (AB)C=a11b11+a12b21a11b12+a12b22a21b11+a22b21a21b12+a22b22c11c12c21c22It follows thatd11= (a11b11+a12b21)c11+ (a11b12+a12b22)c21=a11b11c11+a12b21c11+a11b12c21+a12b22c21d12= (a11b11+a12b21)c12+ (a11b12+a12b22)c22=a11b11c12+a12b21c12+a11b12c22+a12b22c22d21= (a21b11+a22b21)c11+ (a21b12+a22b22)c21=a21b11c11+a22b21c11+a21b12c21+a22b22c21d22= (a21b11+a22b21)c12+ (a21b12+a22b22)c22=a21b11c12+a22b21c12+a21b12c22+a22b22c22If we setE=A(BC) =a11a12a21a22b11c11+b12c21b11c12+b12c22b21c11+b22c21b21c12+b22c22then it follows thate11=a11(b11c11+b12c21) +a12(b21c11+b22c21)=a11b11c11+a11b12c21+a12b21c11+a12b22c21e12=a11(b11c12+b12c22) +a12(b21c12+b22c22)=a11b11c12+a11b12c22+a12b21c12+a12b22c22e21=a21(b11c11+b12c21) +a22(b21c11+b22c21)=a21b11c11+a21b12c21+a22b21c11+a22b22c21e22=a21(b11c12+b12c22) +a22(b21c12+b22c22)=a21b11c12+a21b12c22+a22b21c12+a22b22c22Thusd11=e11d12=e12d21=e21d22=e22and hence(AB)C=D=E=A(BC)9.A2=0010000100000000A3=0001000000000000andA4=O. Ifn >4, thenAn=An−4A4=An−4O=O
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 14 preview imageSection 4•Matrix Algebra910.(a) The matrixCis symmetric sinceCT= (A+B)T=AT+BT=A+B=C(b) The matrixDis symmetric sinceDT= (AA)T=ATAT=A2=D(c) The matrixE=ABis not symmetric sinceET= (AB)T=BTAT=BAand in general,AB6=BA.(d) The matrixFis symmetric sinceFT= (ABA)T=ATBTAT=ABA=F(e) The matrixGis symmetric sinceGT= (AB+BA)T= (AB)T+ (BA)T=BTAT+ATBT=BA+AB=G(f) The matrixHis not symmetric sinceHT= (AB−BA)T= (AB)T−(BA)T=BTAT−ATBT=BA−AB=−H11.(a) The matrixAis symmetric sinceAT= (C+CT)T=CT+ (CT)T=CT+C=A(b) The matrixBis not symmetric sinceBT= (C−CT)T=CT−(CT)T=CT−C=−B(c) The matrixDis symmetric sinceAT= (CTC)T=CT(CT)T=CTC=D(d) The matrixEis symmetric sinceET= (CTC−CCT)T= (CTC)T−(CCT)T=CT(CT)T−(CT)TCT=CTC−CCT=E(e) The matrixFis symmetric sinceFT= ((I+C)(I+CT))T= (I+CT)T(I+C)T= (I+C)(I+CT) =F(e) The matrixGis not symmetric.F= (I+C)(I−CT) =I+C−CT−CCTFT= ((I+C)(I−CT))T= (I−CT)T(I+C)T= (I−C)(I+CT) =I−C+CT−CCTFandFTare not the same. The two middle termsC−CTand−C+CTdo not agree.12.Ifd=a11a22−a21a126= 0, then1da22−a12−a21a11a11a12a21a22=a11a22−a12a21d00a11a22−a12a21d=Ia11a12a21a22[1da22−a12−a21a11]=a11a22−a12a21d00a11a22−a12a21d=I
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 15 preview image10Chapter 1•Matrices and Systems of EquationsTherefore1da22−a12−a21a11=A−113.(b)−352−314.IfAwere nonsingular andAB=A, then it would follow thatA−1AB=A−1Aand hencethatB=I. So ifB6=I, thenAmust be singular.15.SinceA−1A=AA−1=Iit follows from the definition thatA−1is nonsingular and its inverse isA.16.SinceAT(A−1)T= (A−1A)T=I(A−1)TAT= (AA−1)T=Iit follows that(A−1)T= (AT)−117.IfAx=Ayandx6=y, thenAmust be singular, for ifAwere nonsingular, then we couldmultiply byA−1and getA−1Ax=A−1Ayx=y18.Form= 1,(A1)−1=A−1= (A−1)1Assume the result holds in the casem=k, that is,(Ak)−1= (A−1)kIt follows that(A−1)k+1Ak+1=A−1(A−1)kAkA=A−1A=IandAk+1(A−1)k+1=AAk(A−1)kA−1=AA−1=ITherefore(A−1)k+1= (Ak+1)−1and the result follows by mathematical induction.19.IfA2=O, then(I+A)(I−A) =I+A−A+A2=Iand(I−A)(I+A) =I−A+A+A2=IThereforeI−Ais nonsingular and (I−A)−1=I+A.20.IfAk+1=O, then(I+A+· · ·+Ak)(I−A) = (I+A+· · ·+Ak)−(A+A2+· · ·+Ak+1)=I−Ak+1=Iand(I−A)(I+A+· · ·+Ak) = (I+A+· · ·+Ak)−(A+A2+· · ·+Ak+1)=I−Ak+1=I
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Solution Manual for Linear Algebra with Applications, 10th Edition - Page 16 preview imageSection 4•Matrix Algebra11ThereforeI−Ais nonsingular and (I−A)−1=I+A+A2+· · ·+Ak.21.SinceRTR=cosθsinθ−sinθcosθcosθ−sinθsinθcosθ=1001andRRT=cosθ−sinθsinθcosθcosθsinθ−sinθcosθ=1001it follows thatRis nonsingular andR−1=RT22.G2=cos2θ+ sin2θ00cos2θ+ sin2θ=I23.H2= (I−2uuT)2=I−4uuT+ 4uuTuuT=I−4uuT+ 4u(uTu)uT=I−4uuT+ 4uuT=I(sinceuTu= 1)24.In each case, if you square the given matrix, you will end up with the same matrix.25.(a) IfA2=A, then(I−A)2=I−2A+A2=I−2A+A=I−A(b) IfA2=A, then(I−12A)(I+A) =I−12A+A−12A2=I−12A+A−12A=Iand(I+A)(I−12A) =I+A−12A−12A2=I+A−12A−12A=IThereforeI+Ais nonsingular and (I+A)−1=I−12A.26.(a)D2=d2110· · ·00d222· · ·0...00· · ·d2nnSince each diagonal entry ofDis equal to either 0 or 1, it follows thatd2jj=djj, forj= 1, . . . , nand henceD2=D.(b) IfA=XDX−1, thenA2= (XDX−1)(XDX−1) =XD(X−1X)DX−1=XDX−1=A
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