Solution Manual For Linear Algebra With Applications, 8th Edition

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SOLUTIONSMANUALLINEARALGEBRAWITHAPPLICATIONSNINTHEDITIONSteven J. LeonUniversity of Massachusetts, Dartmouth

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ContentsPrefacev1Matrices and Systems of Equations11Systems of Linear Equations12Row Echelon Form23Matrix Arithmetic34Matrix Algebra65Elementary Matrices126Partitioned Matrices17MATLAB Exercises20Chapter Test A22Chapter Test B242Determinants271The Determinant of a Matrix272Properties of Determinants303Additional Topics and Applications33MATLAB Exercises35Chapter Test A35Chapter Test B363Vector Spaces381Definition and Examples382Subspaces423Linear Independence474Basis and Dimension505Change of Basis526Row Space and Column Space52MATLAB Exercises59Chapter Test A60Chapter Test B624Linear Transformations661Definition and Examples662Matrix Representations of Linear Transformations693Similarity71MATLAB Exercise72iii

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ivContentsChapter Test A73Chapter Test B745Orthogonality761The Scalar product inRn762Orthogonal Subspaces783Least Squares Problems814Inner Product Spaces855Orthonormal Sets906The Gram-Schmidt Process987Orthogonal Polynomials100MATLAB Exercises103Chapter Test A104Chapter Test B1056Eigenvalues1091Eigenvalues and Eigenvectors1092Systems of Linear Differential Equations1143Diagonalization1154Hermitian Matrices1235Singular Value Decomposition1306Quadratic Forms1327Positive Definite Matrices1358Nonnegative Matrices138MATLAB Exercises140Chapter Test A144Chapter Test B1457Numerical Linear Algebra1491Floating-Point Numbers1492Gaussian Elimination1503Pivoting Strategies1514Matrix Norms and Condition Numbers1525Orthogonal Transformations1626The Eigenvalue Problem1647Least Squares Problems168MATLAB Exercises171Chapter Test A172Chapter Test B173

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Chapter 1Matrices andSystemsof Equations1SYSTEMS OF LINEAR EQUATIONS2.(d)11111021−210041−20001−3000025.(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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2Chapter 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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Section 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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4Chapter 1•Matrices and Systems of Equations(g)5−10155−148−962.(d)361056103165.(a) 5A=15205510352A+ 3A=6822414+91233621=1520551035(b) 6A=18246612423(2A) = 36822414=1824661242(c)AT=312417(AT)T=312417T=341127=A6.(a)A+B=546051=B+A(b) 3(A+B) = 3546051=15121801533A+ 3B=123186915+390−66−12=1512180153(c) (A+B)T=546051T=504561AT+BT=421365+1−2320−4=5045617.(a) 3(AB) = 35141542016=154245126048(3A)B=63189−6122416=154245126048A(3B) =2163−24612318=154245126048

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Section 3•Matrix Arithmetic5(b) (AB)T=5141542016T=5150144216BTAT=214626−2134=51501442168.(a) (A+B) +C=0517+3121=3638A+ (B+C) =2413+1225=3638(b) (AB)C=−418−2133121=24142011A(BC) =2413−4−184=24142011(c)A(B+C) =24131225=1024717AB+AC=−418−213+14694=1024717(d) (A+B)C=05173121=105178AC+BC=14694+−4−184=1051789.(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=−52a1−14a2.11.The given information implies thatx1=110andx2=011are 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)T

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6Chapter 1•Matrices and Systems of Equations14.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=1215141412121431014121316=43120451203212015.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)TTheith entry ofyis equal to the number of search words in the title of theith book.18.Ifα=a21/a11, then10α1a11a120b=a11a12αa11αa12+b=a11a12a21αa12+bThe 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.

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Section 4•Matrix Algebra72.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=0100andB=1100More generally ifA=abcacbB=dbeb−da−eathenAB=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ThusA2=a2+b2ab+bcab+bcb2+c2IfA2=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+a22b22c11c12c21c22It 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) =a11a12a21a22b11c11+b12c21b11c12+b12c22b21c11+b22c21b21c12+b22c22then it follows thate11=a11(b11c11+b12c21) +a12(b21c11+b22c21)=a11b11c11+a11b12c21+a12b21c11+a12b22c21

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8Chapter 1•Matrices and Systems of Equationse12=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=0010000100000000A3=0001000000000000andA4=O. Ifn >4, thenAn=An−4A4=An−4O=O10.(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

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Section 4•Matrix Algebra9(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, then1da22−a12−a21a11a11a12a21a22=a11a22−a12a21d00a11a22−a12a21d=Ia11a12a21a22[1da22−a12−a21a11]=a11a22−a12a21d00a11a22−a12a21d=ITherefore1da22−a12−a21a11=A−113.(b)−352−314.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=I

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10Chapter 1•Matrices and Systems of EquationsTherefore(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=IThereforeI−Ais nonsingular and (I−A)−1=I+A+A2+· · ·+Ak.21.SinceRTR=cosθsinθ−sinθcosθcosθ−sinθsinθcosθ=1001andRRT=cosθ−sinθsinθcosθcosθsinθ−sinθcosθ=1001it 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· · ·d2nn

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Section 4•Matrix Algebra11Since 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=A27.IfAis an involution, thenA2=Iand it follows thatB2=14 (I+A)2= 14 (I+ 2A+A2) = 14 (2I+ 2A) = 12 (I+A) =BC2=14 (I−A)2= 14 (I−2A+A2) = 14 (2I−2A) = 12 (I−A) =CSoBandCare both idempotent.BC= 14 (I+A)(I−A) = 14 (I+A−A−A2) = 14 (I+A−A−I) =O28.(ATA)T=AT(AT)T=ATA(AAT)T= (AT)TAT=AAT29.LetAandBbe symmetricn×nmatrices. If (AB)T=AB, thenBA=BTAT= (AB)T=ABConversely, ifBA=AB, then(AB)T=BTAT=BA=AB30.(a)BT= (A+AT)T=AT+ (AT)T=AT+A=BCT= (A−AT)T=AT−(AT)T=AT−A=−C(b)A=12(A+AT) +12(A−AT)34.False. For example, ifA=2323,B=1414,x=11thenAx=Bx=55however,A6=B.35.False. For example, ifA=1000andB=0001then it is easy to see that bothAandBmust be singular, however,A+B=I, which isnonsingular.36.True. IfAandBare nonsingular, then their productABmust also be nonsingular. Using theresult from Exercise 23, we have that (AB)Tis nonsingular and ((AB)T)−1= ((AB)−1)T. Itfollows then that((AB)T)−1= ((AB)−1)T= (B−1A−1)T= (A−1)T(B−1)T

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12Chapter 1•Matrices and Systems of Equations5ELEMENTARY MATRICES2.(a)0110,type I(b) The given matrix is not an elementary matrix. Its inverse is given by120013(c)100010−501,type III(d)10001/50001,type II5.(c) SinceC=F B=F EAwhereFandEare elementary matrices, it follows thatCis row equivalent toA.6.(b)E−11=100310001,E−12=100010201,E−13=1000100−11The productL=E−11E−12E−13is lower triangular.L=1003102−117.Acan be reduced to the identity matrix using three row operations2164→2101→2001→1001The elementary matrices corresponding to the three row operations areE1=10−31,E2=1−101,E3=12001SoE3E2E1A=Iand henceA=E−11E−13E−13=103111012001andA−1=E3E2E1.8.(b)10−112405(d)100−2103−21−2120320029.(a)10133422312−3−11−10−23=100010001

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