Transcription of Linear Algebra With Applications
1 With Open TextsLINEAR Algebra with ApplicationsOpen EditionPARTIAL STUDENTSOLUTION MANUALVERSION 2019 REVISION AADAPTABLE | ACCESSIBLE | AFFORDABLEby W. Keith NicholsonCreative Commons License (CC BY-NC-SA)advancing learningChampions of Access to KnowledgeOPEN TEXTONLINEASSESSMENTAll digital forms of access to our high-qualityopen texts are entirely FREE! All content isreviewed for excellence and is wholly adapt-able; custom editions are produced by Lyryxfor those adopting Lyryx assessment. Accessto the original source files is also open to any-one!We have been developing superior online for-mative assessment for more than 15 years. Ourquestions are continuously adapted with thecontent and reviewed for quality and soundpedagogy. To enhance learning, students re-ceive immediate personalized feedback. Stu-dent grade reports and performance statisticsare also to our in-house support team is avail-able 7 days/week to provide prompt resolutionto both student and instructor inquiries.
2 In ad-dition, we work one-on-one with instructors toprovide a comprehensive system, customizedfor their course. This can include adapting thetext, managing multiple sections, and more!Additional instructor resources are also freelyaccessible. Product dependent, these supple-ments include: full sets of adaptable slides andlecture notes, solutions manuals, and multiplechoice question banks with an exam Lyryx learningLinear Algebra with ApplicationsOpen EditionBE A CHAMPION OF OER!Contribute suggestions for improvements, new content, or errata:A new topicA new exampleAn interesting new questionA new or better proof to an existing theoremAny other suggestions to improve the materialContact Lyryx your Keith Nicholson, University of CalgaryLyryx Learning TeamBruce BauslaughPeter ChowNathan FriessStephanie KeyowskiClaude LaflammeMartha LaflammeJennifer MacKenzieTamsyn MurnaghanBogdan SavaRyan YeeLICENSEC reative Commons License (CC BY-NC-SA): This text, including the art and illustrations, are availableunder the Creative Commons license (CC BY-NC-SA), allowinganyone to reuse, revise, remix andredistribute the view a copy of this license, learningLinear Algebra with ApplicationsOpen EditionBase Text Revision HistoryCurrent Revision: Version 2019 Revision A2019 A New Section on Singular Value Decomposition ( ) is included.
3 New Please note that this will impact the numbering of subsequentexamples and theorems in the relevant sections. renamed asMatrix-Vector Multiplication. Minor revisions made throughout, including fixing typos, adding exercises, expanding explanations,and other small B Images have been converted to LaTeX throughout. Text has been converted to LaTeX with minor fixes throughout. Page numbers will differ from 2018 Arevision. Full index has been A Text has been released with a Creative Commons Systems of Linear Solutions and Elementary Operations.. Gaussian Elimination.. Homogeneous equations .. An Application to Network Flows.. An Application to Electrical Networks.. An Application to Chemical Reactions.. 10 Supplementary Exercises: Chapter 1.. 102 matrix matrix Addition, Scalar Multiplication, and Transposition.. matrix -Vector Multiplication.. matrix Multiplication.. matrix Inverses.
4 Elementary Matrices.. matrix Transformations.. LU-factorization.. An Application to Input-Output Economic Models.. An Application to Markov Chains.. 37 Supplementary Exercises: Chapter 2.. 393 Determinants and The Cofactor Expansion.. Determinants and matrix Inverses.. Diagonalization and Eigenvalues.. An Application to Linear Recurrences.. An Application to Systems of Differential equations .. Proof of the Cofactor Expansion Theorem.. 57 Supplementary Exercises: Chapter 3.. 574 Vector Vectors and Lines.. Projections and Planes.. More on the Cross Product.. Linear Operators onR3.. An Application to Computer Graphics.. 73 Supplementary Exercises: Chapter 4.. 735 The Vector Subspaces and Spanning.. Independence and Dimension.. Orthogonality.. Rank of a matrix .. Similarity and Diagonalization.. Best Approximation and Least Squares.. An Application to Correlation and Variance.
5 87 Supplementary Exercises: Chapter 5.. 876 Vector Examples and Basic Properties.. Subspaces and Spanning Sets.. Linear Independence and Dimension.. Finite Dimensional Spaces.. An Application to Polynomials.. An Application to Differential equations .. 101 Supplementary Exercises: Chapter 6.. 1027 Linear Examples and Elementary Properties.. Kernel and Image of a Linear Transformation.. Isomorphisms and Composition.. A Theorem about Differential equations .. More on Linear Recurrences.. 1158 Orthogonal Complements and Projections.. Orthogonal Diagonalization.. Positive Definite Matrices.. QR-Factorization.. Computing Eigenvalues.. Singular Value Decomposition.. Complex Matrices.. An Application to Linear Codes over Finite Fields.. An Application to Quadratic Forms.. An Application to Constrained Optimization.. An Application to Statistical Principal Component Analysis.. 1359 Change of The matrix of a Linear Transformation.
6 Operators and Similarity.. Invariant Subspaces and Direct Sums.. 14610 Inner Product Inner Products and Norms.. Orthogonal Sets of Vectors.. Orthogonal Diagonalization.. Isometries.. An Application to Fourier Approximation.. 16311 Canonical Block Triangular Form.. Jordan Canonical Form.. 167A Complex Numbers169B Proofs175C Mathematical Induction1771. Systems of Linear Solutions and Elementary Operations1. b. Substitute these values ofx1,x2,x3andx4in the equation2x1+5x2+9x3+3x4=2(2s+12t+13)+5(s )+9( s 3t 3)+3(t) = 1x1+2x2+4x3=(2s+12t+13)+2(s)+4( s 3t 3)=1 Hence this is a solution for every value b. The equation is 2x+3y=1. Ifx=stheny=13(1 2s)so this is one form of the generalsolution. Also, ify=tthenx=12(1 3t)gives another Given the equation 4x 2y+0z=1, takey=sandz=tand solve forx:x=14(2s+3). This is thegeneral a. Ifa=0, no solution ifb6=0, infinitely many ifb= Ifa6=0 unique solutionx=b/afor b. The augmented matrix ish1 200 The augmented matrix is 1 1 010 1 10 1 0 12.
7 8. b. A system with this augmented matrix is2x y= 1 3x+2y+z=0y+z=39. 213 4 1i h1210 2 4i h1 210 12i h1 0 30 3,y= 414 5 3i h4 5 33 41i h1 1 43 41i h1 1 40 113i h1 0 170 17,y= b. 2 11 11 2103 0 25 1 2102 11 13 0 25 12100 3 1 10 6 55 "1 2100 113130 0 37# "1 013 230 113130 0 1 73# "1 0 0190 1 01090 0 1 73#. Hencex=19,y=109,z= of Linear Equations11. 25 12816i h3 250036i. The last equation is 0x+0y=36, which has no b. False. The systemx+y=0,x y=0 is consistent, butx=0=yis the only True. If the original system was consistent the final system would also be consistent becauseeach row operation produces a system with the same set of solutions (by Theorem ).16. The substitution gives3(5x 2y )+2( 7x +3y ) =57(5x 2y )+5( 7x +3y ) =1this simplifies tox =5,y =1. Hencex=5x 2y =23 andy= 7x +3y = As in the Hint, multiplying by(x2+2)(2x 1)givesx2 x+3= (ax+b)(2x 1) +c(x2+2).Equating coefficients of powers ofxgives equations 2a+c=1, a+2b= 1, b+2c= this Linear system we finda= 19,b= 59,c= If John gets $xper hour and Joe gets $yper hour, the two situations give 2x+3y= and 3x+2y= Solving givesx=$ andy=$ Gaussian Elimination1.
8 B. No, No; no leading No, Yes; not in reduced form because of the 3 and the top two 1 s in the last No, No; the (reduced) row-echelon form would have two rowsof b."0 1313 210 261 5 0 103 924 1 101 3 13 01# "0 1 3 1 3 2 10 00 1 11 4 30 00513720 000622# "0 1 3 08220 00 111430 00 0 42 13 130 00 0622# "0 1 3 0 8 2 20 00 1 11 4 30 00 0 0 1 10 00 0 3 1 1# "0 1 3 0 8 000 00 1 11 0 10 00 0 3 000 00 0 0 11# "0 1 3 0 0 000 00 1 0 0 10 00 0 1 000 00 0 0 11#3. b. The matrix is already in reduced row-echelon form. The nonleading variables are parameters;x2=r,x4=sandx6= first equation isx1 2x2+2x4+x6=1, whencex1=1+2r 2s second equation isx3+5x4 3x6= 1, whencex3= 1 5s+ third equation isx5+6x6=1, whencex5=1 First carry the matrix to reduced row-echelon form."1 1 2 46201 2 1 1 100 0 10100 0 000# "1 0 4 5510 1 2 1 1 10 0 0 1010 0 0 000# "1 0 4 05 40 1 2 0 1 20 0 0 1010 0 0 000# Gaussian Elimination3 The nonleading variables are parameters;x3=s,x5= first equation isx1+4x3+5x5= 4, whencex1= 4 4s second equation isx2+2x3 x5= 2, whencex2= 2 2s+ third equation isx4= 102 31i h12 12 31i h12 10 73i h1 2 10 1 37i 1 0 170 1 37.
9 Hencex= 17,y= Note that the variables in the second equation are in the wrong 12 62 4i h3 12000i h1 13230 nonleading variabley=tis a parameter; thenx=23+13t=13(t+2).f. Again the order of the variables is reversed in the second 35 232i h2 35007i. There is no solution as the second equation is 0x+0y= b. 23 3 93 4 15 57 2 14 3 4 15 23 3 9 57 2 14 1 1 4 4 23 3 9 57 2 14 1 1 4 401 11 1702 22 34 1 0 15 210 1 11 170 0 00 .Takez=t(the nonleading variable). The equations givex= 21 15t,y= 17 1 2 122 5 311 4 33 1 2 120 1 1 30 2 21 1 2 120 1 1 30 007 .There is no solution as the third equation is 0x+0y+0z= 3 2 1 21 1 35 11 1 1 1 1 353 2 1 2 11 1 1 1 13501 8 170044 1 0 5 120 1 8 170 011 1 0 0 70 1 0 90 0 11 . Hencex= 7,y= 9,z= 12 4102 12511 27 12 4100 5 10 150 12 3 1 2 4100 1 230 000 1 0040 1 230 000 .Hencez=t,x=4,y=3+ b. Label the rows of the augmented matrix asR1,R2andR3, and begin the gaussian algorithm onthe augmented matrix keeping track of the row operations: 12 3 313 551 25 35 R1R2R3 12 5501 280 48 32 R2R2 R1R3 R1At this point observe thatR3 R1= 4(R2 R1), that isR3=5R1 4R2.
10 This means thatequation 3 is 5 times equation 1 minus 4 times equation 2, as isreadily verified. (The solutionisx1=t 11,x2=2t+8 andx3=t.)7. b."1 11 10 1111011 11011110# "1 11 100020002 22002020# "1 11 1001 1100010001010# "1 0 0 000 1 1 100 0 1 000 0 1 00# "1 0 0 000 1 0 100 0 1 000 0 0 00#. Hencex4=t;x1=0,x2= t,x3= of Linear Equationsd."1 12 140 3 1421 2 3501 1 56 3# "1 12 140 3 1420 1 56 40 0 77 7# "1 07 780 0 14 14140 1 56 40 0 77 7# "1 07 780 1 56 40 0 14 14140 0 77 7# "1 00010 1 56 40 01 110 0000# "1 0 0010 1 0110 0 1 110 0 000#.Hencex4=t;x1=1,x2=1 t,x3=1+ 1a25i h1b 10 2 ab5+ 1 Ifab6=2, it continues h1b 10 15+a2 abi 1 0 2 5b2 ab0 15+a2 ab .The unique solution isx= 2 5b2 ab,y=5+a2 2 Ifab=2, it ish1b 10 05+ai. Hence there is no solution ifa6= 5. Ifa= 5, thenb= 25and the matrix ish1 25 10 00i. Theny=t,x= 1+ 1bi h112b2a11i h112b20 1 a21 ab2i h112b20 2 a2 1 Ifa6=2 it continues: 112b20 12 ab2 a 1 0b 12 a0 12 ab2 a.
