Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition

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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition - Page 1 preview image

Solutions ManualElementary LinearAlgebra withApplicationsNinth EditionBernard KolmanDrexel UniversityDavid R. HillTemple University

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ContentsPrefaceiii1Linear Equations and Matrices11.1Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11.2Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21.3Matrix Multiplication. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31.4Algebraic Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71.5Special Types of Matrices and Partitioned Matrices . . . . . . . . . . . . . . . . . . . . . . . .91.6Matrix Transformations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .141.7Computer Graphics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .161.8Correlation Coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .242Solving Linear Systems272.1Echelon Form of a Matrix. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .272.2Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .282.3Elementary Matrices; FindingA−1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .302.4Equivalent Matrices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .322.5LU-Factorization (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .353Determinants373.1Definition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .373.2Properties of Determinants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .373.3Cofactor Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .393.4Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .413.5Other Applications of Determinants. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .434Real Vector Spaces454.1Vectors in the Plane and in 3-Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .454.2Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .474.3Subspaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .484.4Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .514.5Span and Linear Independence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .524.6Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .544.7Homogeneous Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .564.8Coordinates and Isomorphisms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .584.9Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .62

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iiCONTENTSSupplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .64Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .695Inner Product Spaces715.1Standard Inner Product onR2andR3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .715.2Cross Product inR3(Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .745.3Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .775.4Gram-Schmidt Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .815.5Orthogonal Complements. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .845.6Least Squares (Optional). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .85Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .86Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .906Linear Transformations and Matrices936.1Definition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .936.2Kernel and Range of a Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . .966.3Matrix of a Linear Transformation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .976.4Vector Space of Matrices and Vector Space of Linear Transformations (Optional) . . . . . . .996.5Similarity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1026.6Introduction to Homogeneous Coordinates (Optional). . . . . . . . . . . . . . . . . . . . . .103Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .105Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1067Eigenvalues and Eigenvectors1097.1Eigenvalues and Eigenvectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1097.2Diagonalization and Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1157.3Diagonalization of Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .120Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .123Chapter Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1268Applications of Eigenvalues and Eigenvectors (Optional)1298.1Stable Age Distribution in a Population; Markov Processes. . . . . . . . . . . . . . . . . . .1298.2Spectral Decomposition and Singular Value Decomposition. . . . . . . . . . . . . . . . . . .1308.3Dominant Eigenvalue and Principal Component Analysis. . . . . . . . . . . . . . . . . . . .1308.4Differential Equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1318.5Dynamical Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1328.6Real Quadratic Forms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1338.7Conic Sections. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1348.8Quadric Surfaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13510 MATLAB Exercises137Appendix B Complex Numbers163B.1Complex Numbers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .163B.2Complex Numbers in Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .165

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Chapter 1Linear Equations and MatricesSection 1.1, p. 82.x= 1,y= 2,z=−2.4. No solution.6.x= 13 + 10t,y=−8−8t,tany real number.8. Inconsistent; no solution.10.x= 2,y=−1.12. No solution.14.x=−1,y= 2,z=−2.16. (a) For example:s= 0,t= 0 is one answer.(b) For example:s= 3,t= 4 is one answer.(c)s=t2.18. Yes. The trivial solution is always a solution to a homogeneous system.20.x= 1,y= 1,z= 4.22.r=−3.24. Ifx1=s1,x2=s2,. . .,xn=snsatisfy each equation of (2) in the original order, then thosesame numbers satisfy each equation of (2) when the equations are listed with one of the original onesinterchanged, and conversely.25. Ifx1=s1,x2=s2,. . .,xn=snis a solution to (2), then thepth andqth equations are satisfied.That is,ap1s1+· · ·+apnsn=bpaq1s1+· · ·+aqnsn=bq.Thus, for any real numberr,(ap1+raq1)s1+· · ·+ (apn+raqn)sn=bp+rbq.Then if theqth equation in (2) is replaced by the preceding equation, the valuesx1=s1,x2=s2,. . .,xn=snare a solution to the new linear system since they satisfy each of the equations.

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2Chapter 126.(a) A unique point.(b) There are infinitely many points.(c) No points simultaneously lie in all three planes.28. No points of intersection:C1C2C2C1One point of intersection:C1C2Two points of intersection:C1C2Infinitely many points of intersection:C1C2=30. 20 tons of low-sulfur fuel, 20 tons of high-sulfur fuel.32. 3.2 ounces of food A, 4.2 ounces of food B, and 2 ounces of food C.34.(a)p(1) =a(1)2+b(1) +c=a+b+c=−5p(−1) =a(−1)2+b(−1) +c=a−b+c= 1p(2) =a(2)2+b(2) +c= 4a+ 2b+c= 7.(b)a= 5,b=−3,c=−7.Section 1.2, p. 192. (a)A=⎡⎢⎢⎢⎢⎢⎣0100110111010000100011000⎤⎥⎥⎥⎥⎥⎦(b)A=⎡⎢⎢⎢⎢⎢⎣0111110100110101010010000⎤⎥⎥⎥⎥⎥⎦.4.a= 3,b= 1,c= 8,d=−2.6.(a)C+E=E+C=⎡⎣5−58429534⎤⎦.(b) Impossible.(c)[7−701].(d)⎡⎣−93−9−12−3−15−6−3−9⎤⎦.(e)⎡⎣010−98−1−2−5−43⎤⎦.(f) Impossible.8.(a)AT=⎡⎣122134⎤⎦, (AT)T=[123214].(b)⎡⎣545−523894⎤⎦.(c)[−6101117].

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Section 1.33(d)[0−440].(e)⎡⎣3463910⎤⎦.(f)[172−166].10. Yes: 2[1001]+ 1[1000]=[3002].12.⎡⎣λ−1−2−3−6λ+ 2−3−5−2λ−4⎤⎦.14. Because the edges can be traversed in either direction.16. Letx=⎡⎢⎢⎢⎣x1x2...xn⎤⎥⎥⎥⎦be ann-vector. Thenx+0=⎡⎢⎢⎢⎣x1x2...xn⎤⎥⎥⎥⎦+⎡⎢⎢⎢⎣00...0⎤⎥⎥⎥⎦=⎡⎢⎢⎢⎣x1+ 0x2+ 0...xn+ 0⎤⎥⎥⎥⎦=⎡⎢⎢⎢⎣x1x2...xn⎤⎥⎥⎥⎦=x.18.n∑i=1m∑j=1aij= (a11+a12+· · ·+a1m) + (a21+a22+· · ·+a2m) +· · ·+ (an1+an2+· · ·+anm)= (a11+a21+· · ·+an1) + (a12+a22+· · ·+an2) +· · ·+ (a1m+a2m+· · ·+anm)=m∑j=1n∑i=1aij.19.(a) True.n∑i=1(ai+ 1) =n∑i=1ai+n∑i=11 =n∑i=1ai+n.(b) True.n∑i=1⎛⎝m∑j=11⎞⎠=n∑i=1m=mn.(c) True.⎡⎣n∑i=1ai⎤⎦⎡⎣m∑j=1bj⎤⎦=a1m∑j=1bj+a2m∑j=1bj+· · ·+anm∑j=1bj= (a1+a2+· · ·+an)m∑j=1bj=n∑i=1aim∑j=1bj=m∑j=1(n∑i=1aibj)20. “new salaries” =u+.08u= 1.08u.Section 1.3, p. 302.(a) 4.(b) 0.(c) 1.(d) 1.4.x= 5.

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4Chapter 16.x=±√2,y=±3.8.x=±5.10.x=65,y=125.12.(a) Impossible.(b)⎡⎣0−111251719022⎤⎦.(c)⎡⎣15−71423−52913−117⎤⎦.(d)⎡⎣881413139⎤⎦.(e) Impossible.14.(a)[58126613].(b) Same as (a).(c)[2883834441].(d) Same as (c).(e)[28321618]; same.(f)[−16−8−26−300−31].16.(a) 1.(b)−6.(c)[−301].(d)⎡⎣−142−2843−12−6⎤⎦.(e) 10.(f)⎡⎣90−3000−301⎤⎦.(g) Impossible.18.DI2=I2D=D.20.[0000].22.(a)⎡⎢⎢⎣114013⎤⎥⎥⎦.(b)⎡⎢⎢⎣018313⎤⎥⎥⎦.24. col1(AB) = 1⎡⎣123⎤⎦+ 3⎡⎣−240⎤⎦+ 2⎡⎣−13−2⎤⎦; col2(AB) =−1⎡⎣123⎤⎦+ 2⎡⎣−240⎤⎦+ 4⎡⎣−13−2⎤⎦.26. (a)−5.(b)BAT28. LetA=[aij]bem×pandB=[bij]bep×n.(a) Let theith row ofAconsist entirely of zeros, so thataik= 0 fork= 1,2, . . . , p. Then the (i, j)entry inABisp∑k=1aikbkj= 0forj= 1,2, . . . , n.(b) Let thejth column ofAconsist entirely of zeros, so thatakj= 0 fork= 1,2, . . . , m. Then the(i, j) entry inBAism∑k=1bikakj= 0fori= 1,2, . . . , m.30.(a)⎡⎢⎢⎣23−31130203230−4000111⎤⎥⎥⎦.(b)⎡⎢⎢⎣23−31130203230−4000111⎤⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣x1x2x3x4x5⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎣7−235⎤⎥⎥⎦.

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Section 1.35(c)⎡⎢⎢⎣23−311730203−2230−403001115⎤⎥⎥⎦32.[−231−5] [x1x2]=[54].34.(a)2x1+x2+ 3x3+ 4x4= 03x1−x2+ 2x3= 3−2x1+x2−4x3+ 3x4= 2(b) same as (a).36.(a)x1[31]+x2[2−1]+x3[14]=[4−2].(b)x1⎡⎣−123⎤⎦+x2⎡⎣1−11⎤⎦=⎡⎣3−21⎤⎦.38.(a)[120253] ⎡⎣x1x2x3⎤⎦=[11].(b)⎡⎣121112202⎤⎦⎡⎣x1x2x3⎤⎦=⎡⎣000⎤⎦.39. We haveu·v=n∑i=1uivi=[u1u2· · ·un]⎡⎢⎢⎢⎣v1v2...vn⎤⎥⎥⎥⎦=uTv.40. Possible answer:⎡⎣100200300⎤⎦.42.(a) Can say nothing.(b) Can say nothing.43.(a) Tr(cA) =n∑i=1caii=cn∑i=1aii=cTr(A).(b) Tr(A+B) =n∑i=1(aii+bii) =n∑i=1aii+n∑i=1bii= Tr(A) + Tr(B).(c) LetAB=C=[cij]. ThenTr(AB) = Tr(C) =n∑i=1cii=n∑i=1n∑k=1aikbki=n∑k=1n∑i=1bkiaik= Tr(BA).(d) SinceaTii=aii, Tr(AT) =n∑i=1aTii=n∑i=1aii= Tr(A).(e) LetATA=B=[bij]. Thenbii=n∑j=1aTijaji=n∑j=1a2ji=⇒Tr(B) = Tr(ATA) =n∑i=1bii=n∑i=1n∑j=1a2ij≥0.Hence, Tr(ATA)≥0.

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6Chapter 144.(a) 4.(b) 1.(c) 3.45. We have Tr(AB−BA) = Tr(AB)−Tr(BA) = 0, while Tr([1001])= 2.46.(a) LetA=[aij]andB=[bij]bem×nandn×p, respectively. Thenbj=⎡⎢⎢⎢⎣b1jb2j...bnj⎤⎥⎥⎥⎦and theithentry ofAbjisn∑k=1aikbkj, which is exactly the (i, j) entry ofAB.(b) Theith row ofABis[∑kaikbk1∑kaikbk2· · ·∑kaikbkn]. Sinceai=[ai1ai2· · ·ain],we haveaib=[∑kaikbk1∑kaikbk2· · ·∑kaikbkn].This is the same as theith row ofAb.47. LetA=[aij]andB=[bij]bem×nandn×p, respectively. Then thejth column ofABis(AB)j=⎡⎢⎣a11b1j+· · ·+a1nbnj...am1b1j+· · ·+amnbnj⎤⎥⎦=b1j⎡⎢⎣a11...am1⎤⎥⎦+· · ·+bnj⎡⎢⎣a1n...amn⎤⎥⎦=b1jCol1(A) +· · ·+bnjColn(A).Thus thejth column ofABis a linear combination of the columns ofAwith coefficients the entries inbj.48. The value of the inventory of the four types of items.50.(a) row1(A)·col1(B) = 80(20) + 120(10) = 2800 grams of protein consumed daily by the males.(b) row2(A)·col2(B) = 100(20) + 200(20) = 6000 grams of fat consumed daily by the females.51.(a) No. Ifx= (x1, x2, . . . , xn), thenx·x=x21+x22+· · ·+x2n≥0.(b)x=0.52. Leta= (a1, a2, . . . , an),b= (b1, b2, . . . , bn), andc= (c1, c2, . . . , cn). Then(a)a·b=n∑i=1aibiandb·a=n∑i=1biai, soa·b=b·a.(b) (a+b)·c=n∑i=1(ai+bi)ci=n∑i=1aici+n∑i=1bici=a·c+b·c.(c) (ka)·b=n∑i=1(kai)bi=kn∑i=1aibi=k(a·b).

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Section 1.4753. Thei,ith element of the matrixAATisn∑k=1aikaTki=n∑k=1aikaik=n∑k=1(aik)2.Thus ifAAT=O, then each sum of squaresn∑k=1(aik)2equals zero, which impliesaik= 0 for eachiandk. ThusA=O.54.AC=[1722218323].CAcannot be computed.55.BTBwill be 6×6 whileBBTis 1×1.Section 1.4, p. 401. LetA=[aij],B=[bij],C=[cij].Then the (i, j) entry ofA+ (B+C) isaij+ (bij+cij) andthat of (A+B) +Cis (aij+bij) +cij. By the associative law for addition of real numbers, these twoentries are equal.2. ForA=[aij], letB=[−aij].4. LetA=[aij],B=[bij],C=[cij]. Then the (i, j) entry of (A+B)Cisn∑k=1(aik+bik)ckjand that ofAC+BCisn∑k=1aikckj+n∑k=1bikckj. By the distributive and additive associative laws for real numbers,these two expressions for the (i, j) entry are equal.6. LetA=[aij], whereaii=kandaij= 0 ifi=j, and letB=[bij]. Then, ifi=j, the (i, j) entry ofABisn∑s=1aisbsj=kbij, while ifi=j, the (i, i) entry ofABisn∑s=1aisbsi=kbii. ThereforeAB=kB.7. LetA=[aij]andC=[c1c2· · ·cm].ThenCAis a 1×nmatrix whoseith entry isn∑j=1cjaij.SinceAj=⎡⎢⎢⎢⎣a1ja2j...amj⎤⎥⎥⎥⎦, theith entry ofn∑j=1cjAjism∑j=1cjaij.8. (a)[cos 2θsin 2θ−sin 2θcos 2θ].(b)[cos 3θsin 3θ−sin 3θcos 3θ].(c)[coskθsinkθ−sinkθcoskθ].(d) The result is true forp= 2 and 3 as shown in parts (a) and (b). Assume that it is true forp=k.ThenAk+1=AkA=[coskθsinkθ−sinkθcoskθ] [cosθsinθ−sinθcosθ]=[coskθcosθ−sinkθsinθcoskθsinθ+ sinkθcosθ−sinkθcosθ−coskθsinθcoskθcosθ−sinkθsinθ]=[cos(k+ 1)θsin(k+ 1)θ−sin(k+ 1)θcos(k+ 1)θ].Hence, it is true for all positive integersk.

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8Chapter 110. Possible answers:A=[1001];A=[0110];A=⎡⎣1√21√21√2−1√2⎤⎦.12. Possible answers:A=[11−1−1];A=[0000];A=[0100].13. LetA=[aij]. The (i, j) entry ofr(sA) isr(saij), which equals (rs)aijands(raij).14. LetA=[aij]. The (i, j) entry of (r+s)Ais (r+s)aij, which equalsraij+saij, the (i, j) entry ofrA+sA.16. LetA=[aij], andB=[bij]. Thenr(aij+bij) =raij+rbij.18. LetA=[aij]andB=[bij]. The (i, j) entry ofA(rB) isn∑k=1aik(rbkj), which equalsrn∑k=1aikbkj, the(i, j) entry ofr(AB).20.16A,k=16.22. 3.24. IfAx=rxandy=sx, thenAy=A(sx) =s(Ax) =s(rx) =r(sx) =ry.26. The (i, j) entry of (AT)Tis the (j, i) entry ofAT, which is the (i, j) entry ofA.27. (b) The (i, j) entry of (A+B)Tis the (j, i) entry of[aij+bij], which is to say,aji+bji.(d) LetA=[aij]and letbij=aji. Then the (i, j) entry of (cA)Tis the (j, i) entry of[caij], whichis to say,cbij.28. (A+B)T=⎡⎣505212⎤⎦, (rA)T=⎡⎣−4−8−12−4−812⎤⎦.30. (a)⎡⎣−3417−51⎤⎦.(b)⎡⎣−3417−51⎤⎦.(c)BTCis a real number (a 1×1 matrix).32. Possible answers:A=[1−300];B=[12231];C=[−1201].A=[2030];B=[0010];C=[0001].33. The (i, j) entry ofcAiscaij, which is 0 for alliandjonly ifc= 0 oraij= 0 for alliandj.34. LetA=[abcd]be such thatAB=BAfor any 2×2 matrixB. Then in particular,[abcd] [1000]=[1000] [abcd][a0c0]=[ab00]sob=c= 0,A=[a00d].

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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition - Page 13 preview image

Section 1.59Also[a00d] [1100]=[1100] [a00d][aa00]=[ad00],which implies thata=d. ThusA=[a00a]for some numbera.35. We have(A−B)T= (A+ (−1)B)T=AT+ ((−1)B)T=AT+ (−1)BT=AT−BTby Theorem 1.4(d)).36. (a)A(x1+x2) =Ax1+Ax2=0+0=0.(b)A(x1−x2) =Ax1−Ax2=0−0=0.(c)A(rx1) =r(Ax1) =r0=0.(d)A(rx1+sx2) =r(Ax1) +s(Ax2) =r0+s0=0.37. We verify thatx3is also a solution:Ax3=A(rx1+sx2) =rAx1+sAx2=rb+sb= (r+s)b=b.38. IfAx1=bandAx2=b, thenA(x1−x2) =Ax1−Ax2=b−b=0.Section 1.5, p. 521. (a) LetIm=[dij]sodij= 1 ifi=jand 0 otherwise. Then the (i, j) entry ofImAism∑k=1dikakj=diiaij(since all otherd’s = 0)=aij(sincedii= 1).2. We prove that the product of two upper triangular matrices is upper triangular: LetA=[aij]withaij= 0 fori > j; letB=[bij]withbij= 0 fori > j. ThenAB=[cij]wherecij=n∑k=1aikbkj. Fori > j, and each 1≤k≤n, eitheri > k(and soaik= 0) or elsek≥i > j(sobkj= 0). Thus everyterm in the sum forcijis 0 and socij= 0. Hence[cij]is upper triangular.3. LetA=[aij]andB=[bij], where bothaij= 0 andbij= 0 ifi=j. Then ifAB=C=[cij], wehavecij=n∑k=1aikbkj= 0 ifi=j.4.A+B=⎡⎣9−110−27003⎤⎦andAB=⎡⎣18−5110−8−7000⎤⎦.5. All diagonal matrices.

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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition - Page 14 preview image

10Chapter 16.(a)[7−2−310](b)[−9−112213](c)[20−20476]8.ApAq= (A·A· · ·A)︸︷︷︸pfactors(A·A· · ·A)︸︷︷︸qfactors︸︷︷︸p+qfactors=Ap+q;(Ap)q=ApApAp· · ·Ap︸︷︷︸qfactors=Aqsummands︷︸︸︷p+p+· · ·+p=Apq.9. We are given thatAB=BA.Forp= 2, (AB)2= (AB)(AB) =A(BA)B=A(AB)B=A2B2.Assume that forp=k, (AB)k=AkBk. Then(AB)k+1= (AB)k(AB) =AkBk·A·B=Ak(Bk−1AB)B=Ak(Bk−2AB2)B=· · ·=Ak+1Bk+1.Thus the result is true forp=k+ 1. Hence it is true for all positive integersp. Forp= 0, (AB)0=In=A0B0.10. Forp= 0, (cA)0=In= 1·In=c0·A0. Forp= 1,cA=cA. Assume the result is true forp=k:(cA)k=ckAk, then fork+ 1:(cA)k+1= (cA)k(cA) =ckAk·cA=ck(Akc)A=ck(cAk)A= (ckc)(AkA) =ck+1Ak+1.11. True forp= 0: (AT)0=In=ITn= (A0)T. Assume true forp=n. Then(AT)n+1= (AT)nAT= (An)TAT= (AAn)T= (An+1)T.12. True forp= 0: (A0)−1=I−1n=In. Assume true forp=n. Then(An+1)−1= (AnA)−1=A−1(An)−1=A−1(A−1)n= (A−1)n+1.13.(1kA−1)(kA) =(1k·k)A−1A=Inand (kA)(1kA−1)=(k·1k)AA−1=In. Hence, (kA)−1=1kA−1fork= 0.14. (a) LetA=kIn. ThenAT= (kIn)T=kITn=kIn=A.(b) Ifk= 0, thenA=kIn= 0In=O, which is singular. Ifk= 0, thenA−1= (kA)−1=1kA−1, soAis nonsingular.(c) No, the entries on the main diagonal do not have to be the same.16. Possible answers:[ab0a]. Infinitely many.17. The result is false. LetA=[1234]. ThenAAT=[5111125]andATA=[10141420].18. (a)Ais symmetric if and only ifAT=A, or if and only ifaij=aTij=aji.(b)Ais skew symmetric if and only ifAT=−A, or if and only ifaTij=aji=−aij.(c)aii=−aii, soaii= 0.19. SinceAis symmetric,AT=Aand so (AT)T=AT.20. The zero matrix.21. (AAT)T= (AT)TAT=AAT.22. (a) (A+AT)T=AT+ (AT)T=AT+A=A+AT.

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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition - Page 15 preview image

Section 1.511(b) (A−AT)T=AT−(AT)T=AT−A=−(A−AT).23. (Ak)T= (AT)k=Ak.24. (a) (A+B)T=AT+BT=A+B.(b) IfABis symmetric, then (AB)T=AB, but (AB)T=BTAT=BA, soAB=BA. Conversely, ifAB=BA, then (AB)T=BTAT=BA=AB, soABis symmetric.25.(a) LetA=[aij]be upper triangular, so thataij= 0 fori > j. SinceAT=[aTij], whereaTij=aji,we haveaTij= 0 forj > i, oraTij= 0 fori < j. HenceATis lower triangular.(b) Proof is similar to that for (a).26. Skew symmetric.To show this, letAbe a skew symmetric matrix.ThenAT=−A.Therefore(AT)T=A=−AT. HenceATis skew symmetric.27. IfAis skew symmetric,AT=−A. Thusaii=−aii, soaii= 0.28. Suppose thatAis skew symmetric, soAT=−A.Then (Ak)T= (AT)k= (−A)k=−Akifkis apositive odd integer, soAkis skew symmetric.29. LetS=(12)(A+AT) andK=(12)(A−AT).ThenSis symmetric andKis skew symmetric, byExercise 18. ThusS+K=(12)(A+AT+A−AT) =(12)(2A) =A.Conversely, supposeA=S+Kis any decomposition ofAinto the sum of a symmetric and skewsymmetric matrix. ThenAT= (S+K)T=ST+KT=S−KA+AT= (S+K) + (S−K) = 2S,S=(12)(A+AT),A−AT= (S+K)−(S−K) = 2K,K=(12)(A−AT)30.S= 12⎡⎣2737123336⎤⎦andK= 12⎡⎣0−1−71017−10⎤⎦.31. Form[2346] [wxyz]=[1001]. Since the linear systems2w+ 3y= 14w+ 6y= 0and2x+ 3z= 04x+ 6z= 1have no solutions, we conclude that the given matrix is singular.32.D−1=⎡⎢⎢⎢⎣14000−1200013⎤⎥⎥⎥⎦.34.A=⎡⎣−12122−1⎤⎦.36. (a)[1213] [46]=[1622].(b)[3853].

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Solution Manual for Elementary Linear Algebra with Applications (Classic Version), 9th Edition - Page 16 preview image

12Chapter 138.[−9−6].40.[89].42. Possible answer:[1000]+[0001]=[1001].43. Possible answer:[1234]+[−1−234]=[0068].44. The conclusion of the corollary is true forr= 2, by Theorem 1.6.Supposer≥3 and that theconclusion is true for a sequence ofr−1 matrices. Then(A1A2· · ·Ar)−1= [(A1A2· · ·Ar−1)Ar]−1=A−1r(A1A2· · ·Ar−1)−1=A−1rA−1r−1· · ·A−12A−11.45. We haveA−1A=In=AA−1and since inverses are unique, we conclude that (A−1)−1=A.46. Assume thatAis nonsingular, so that there exists ann×nmatrixBsuch thatAB=In. Exercise 28in Section 1.3 implies thatABhas a row consisting entirely of zeros. Hence, we cannot haveAB=In.47. LetA=⎡⎢⎢⎢⎣a1100· · ·00a220· · ·0...00· · ·· · ·ann⎤⎥⎥⎥⎦,whereaii= 0 fori= 1,2, . . . , n. ThenA−1=⎡⎢⎢⎢⎢⎢⎢⎣1a1100· · ·001a220· · ·0...00· · ·· · ·1ann⎤⎥⎥⎥⎥⎥⎥⎦as can be verified by computingAA−1=A−1A=In.48.A4=⎡⎣1600081000625⎤⎦.49.Ap=⎡⎢⎢⎢⎣ap1100· · ·00ap220· · ·0...00· · ·· · ·apnn⎤⎥⎥⎥⎦.50. Multiply both sides of the equation byA−1.51. Multiply both sides byA−1.

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