WIXLIB

2Systems of Linear Equationsh3 12 28516i h30 20536i. The last equation is 0x 0y 36, which has no solution.11.b.14.b. False. The system x y 0, x y 0 is consistent, but x 0 y is the only solution.d. 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 1.1.1).16. The substitution gives3(5x′ 2y′ ) 2( 7x′ 3y′ ) 57(5x′ 2y′ ) 5( 7x′ 3y′ ) 1this simplifies to x′ 5, y′ 1. Hence x 5x′ 2y′ 23 and y 7x′ 3y′ 32.17. As in the Hint, multiplying by (x2 2)(2x 1) gives x2 x 3 (ax b)(2x 1) c(x2 2).Equating coefficients of powers of x gives equations 2a c 1, a 2b 1, b 2c 3.Solving this linear system we find a 19 , b 59 , c 119.19. If John gets x per hour and Joe gets y per hour, the two situations give 2x 3y 24.6 and 3x 2y 23.9. Solving gives x 4.50 and y 5.20.1.2 Gaussian Elimination1.b. No, No; no leading 1.d. No, Yes; not in reduced form because of the 3 and the top two 1’s in the last column.2.3.f. No, No; the (reduced) row-echelon form would have two rows of zeros." 0 1 3 1 3 2 1 # " 0 1 3 1 3 2 1 #000b. " " 2310000000061 5 0 1 00 924 1 1 3 13 01# 01 3 082200 11143 00 0 42 13 1300 0622# " 01 3 080000 1 11 0 1 0000 030000 00110000"10000000 150 300001000001000 3000 11 4 31372622#082 21 11 4 3001 1031 1#0 000 0 11 000 11b. The matrix is already in reduced row-echelon form. The nonleading variables are parameters;x2 r, x4 s and x6 t.The first equation is x1 2x2 2x4 x6 1, whence x1 1 2r 2s t.The second equation is x3 5x4 3x6 1, whence x3 1 5s 3t.The third equation is x5 6x6 1, whence x5 1 6t.d. First carry the matrix to reduced row-echelon form." 1 1 2 4 6 2 # " 1 0 4 5 5000100200110 100 110 000100200110 1001 110# "10000100420000105 100 4 210#

1.2. Gaussian Elimination4.The nonleading variables are parameters; x3 s, x5 t.The first equation is x1 4x3 5x5 4, whence x1 4 4s 5t.The second equation is x2 2x3 x5 2, whence x2 2 2s t.The third equation is x4 1.hi hi hi hi 11 2 13 1 012 112 1b. 2 3 1 2 3 1 0 7 3 0 1 3 0701Hence x 17 , y 37 . 17 37 3.d. hNote that the variablesarei h in the secondi h equationi in the wrong order.21 313 1 23 123 0 0 0 0 0 0 . 62 4The nonleading variable y t is a parameter; then x 23 13 t 31 (t 2).f. hAgain the orderi ofh the variablesi is reversed in the second equation.2 3 52 3 5 0 0 7 . There is no solution as the second equation is 0x 0y 7. 23 25.b. 23 5 93 43 3 4 15 2 37 2 147 5 1 14 41 001 11 17 0 102 22 340 0513 92 1415 2111 1700 . 1 2 5 137 4 9 14432 Take z t (the nonleading variable). The equations give x 21 15t, y 17 11t. 1 2 1 2221 2 11 2 1d. 2 5 3 1 0 1 1 3 0 1 1 3 .1 3430 2210007There is no solution as the third equation is 0x 0y 0z 7. 3 2 1 21 1 351 1351 1 353 2 1 2f. 0 1 8 174 11 1 1 11 1 1004 1 0 5 121 0 0 7 0 1 8 17 0 1 0 9 . Hence x 7, y 9, z 1.0 0110 0 11 1012 4 1012 41 2 4 101 0h. 2 1 2 5 0 5 10 15 0 1 2 3 0 111 27 102Hence z t, x 4, y 3 2t.6.000000430 .b. Label the rows of the augmented matrix as R1 , R2 and R3 , and begin the gaussian algorithm onthe augmented matrix keeping track of the row operations: R112 5R212 3 3513 55R2 01 28R2 R117. 30 20b. 25 35R30 48 32R3 R1At this point observe that R3 R1 4(R2 R1 ), that is R3 5R1 4R2 . This means thatequation 3 is 5 times equation 1 minus 4 times equation 2, as is readily verified. (The solutionis x1 t 11, x2 2t 8 and x3 t.)" 1 1 1 1 0 # " 1 1 1 1 0 # " 1 1 1 1 0 # 111 "1000111 111001 101011110100000#0000 "00010000100020 0 2 22 0202 0#0 0 00 1 0. Hence1 0 00 0 0000101 110101000x4 t; x1 0, x2 t, x3 0.

4Systems of Linear Equationsd.1011" 8."132110002 14 142 350 56 307 71 56014 140 77# 8 414 7"#1000 13101000" 14670 5102 1 5 7010042 4 706 10#1 410Hence x4 t; x1 1, x2 1 t, x3 1 t.hi hib. a1 2b 15 10 2 bab 5 1 a .Case 1 If ab 6 2, it continues h10 1b1 5 a2 abi"1000# 00101000"1001 78 14146 47 700 101 11 1 100 0714 5 70100 2 5b2 ab5 a2 ab ##.5 aThe unique solution is x 2 5b2 abi , y 2 ab .hCase 2 If ab 2, it is 10 0b 5 1 a . Hence there is no solution if a 6 5. If a 5, thenhi1 2 15b 2andthematrixis. Then y t, x 1 25 t.5000hi hi hii hb1b1b111 12a 1 122222d. 2 1 b a 1 1 0 1 a 1 ab 0 2 a 2 ab .22 1bb 11 21022 aCase 1 If a 6 2 it continues: .2 ab2 ab0The unique solution: x Case 2 If a 2 the matrix ish1If b 1 the matrix is 09.b. 201100120012 13 1 113abc ca 2cb 102 021100b 12 a ,hy 0121200 02 a12 a2 ab2 a .1210 1 c3 b 1 a0 111011i. Hence there is no solution if b 6 1.b2i 2(1 b), so y t, x 21 12 t 12 (1 t). 100021ca 2cb 2a 4c 131 cb a 2c1 0 0 b 2a 5c0 1 0 3a b 6c0 0 1 b 2a 4c .Hence, for any values of a, b and c there is a unique solution x 2a b 5c, y 3a b 6c,and z 2a b 4c. 1 a 0 01a0 01 0 ab0b0 .d. 0 1 b 0 0 1 b 0 0 10c100 ac100Case 1 If abc 6 1, it continues: 10000101 abc0 abb1 000 100010001000 .Hence we have the unique solution x 0, y 0, z 0. 1 0 ab 001b0Case 2 If abc 1, the matrix is, so z t, x abt, y bt.0000Note: It is impossible that there is no solution here: x y z 0 always works. 1a 111a 111a 110f. 1 a 2 1 1 0 2(a 1) 0 0 a 1 0 0 .22a 210Case 1 If a 1 the matrix is2(a 1) 100100a 101 110 10 1000100a0100 10 1 ,so y t, x t, z 1.Case 2 If a 0 the last equation is 0x 0y 0z 1, so there is no solution.

1.2. Gaussian EliminationCase 3 If a 6 1 and a 6 is a unique solution: 0, there 1 a 111a 111 000100 a 1000 1 1010.b.h200 1a00 a11 a10 1a00100 Hence x 1 a1 , y 0, z 1a .ii h31 12 211 1 32 0 0 0 0 ; rank is 1.00 0 5.d. It is in row-echelon form; rank is 3. 0 0 10 0 1f. 0 0 1 0 0 0 ; rank is 1.011.b.d.f. 01 23 53 4731231 1 2 11131rank is 3. 11211 a2 a 00 13 5 25 1226 aIf a 0 we getIf a 2 we geta204 1001000 100120If a 6 0, a 6 2, we get12.b. False. A d. False. A 100010110100010110 41213 1 1 21 2a210 a20 2 a 4 2a20 1 1 2 0 0 0 1 2 ; rank 0 0 0 0 1 1 2 4 0 1 0 2 ; rank0 0 0 0 12a21 1a0a20 1 0 00 2 a 4 a2202200 100004440100 311 1 12 1 471005 2 1 4 1122 100 1 a a 1101003 84 1a0 1105 1744110 202 a ; rank is 2. a2a24 a2 110100 3 815 171 ;. 2. 2.201a2a2 a ; rank 3.f. False. The system 2x y 0, 4x 2y 0 is consistent, but the system 2x y 1, 4x 2y 1 is not consistent.14.h. True. A has 3 rows so there can be at most 3 leading 1’s. Hence the rank of A is at most 3. 1 a b c1ab cb. We begin the row reduction 1 b c a 0 b a a b . Now one of b a and c 1a bc0c aa c00a is nonzero (by hypothesis) so that row provides the second leading 1 (its row becomes0 1 1 ). Hence further row operations give 1 a b c1 0 b c a 1 0 1 1 0 10000which has the given form.16.b. Substituting the coordinates of the three points in the equation gives1 1 a b c 025 9 5a 3b c 09 9 3a 3b c 0a b c 25a 3b c 343a 3b c 18

6Systems of Linear Equations 153 "1 33100010 2 341811 11212 53 61 # 100"1 80100010001 2 24241 4 4 26 6# 100110 23 61121 .Hence a 2, b 6, c 6, so the equation is x2 y2 2x 6y 6 0.18. Let a, b and c denote the fractions of the student population in Clubs A, B and C respectively. The42of those in Club A stayed; 10of those in Club B go tonew students in Club A arrived as follows: 102A, and 10 of those in C go to A. Hencea 22410 a 10 b 10 cSimilarly, looking at students in Club B and C.b 17210 a 10 b 10 cc 51610 a 10 b 10 cHence 6a 2b 2c 0a 3b 2c 05a b 4c 0 6152 3100022 4 3 1616100214 14000 3101002 870 000 "100010 85 870#000.Thus the solution is a 58 t, b 78 t, c t. However a b c 1 (because every student belongs to578exactly one club) which gives t 52 . Hence a 20, b 20, c 20.1.3 Homogeneous Equations1.b. False. A False. A f.2.b. 11223316a000hh 10011100100100i. 10021 1i.d. False. A h. False. A 15a 2000 100010 95a 3000h 10011110100010000 . .Hence there is a nontrivial solution when a 3: x 9t, y 5t, z t. a 11 01 1 1 011 10d. 1 1 1 0 a 1 1 0 0 1 a 1 a 0 .11a011a000a 1i.0Hence if a 6 1 and a 6 1, there is a unique, trivial solution. The other cases are as follows:

1.3. Homogeneous Equationsa 1: a 1 :3.100 100100 1 02 02 01 1 020 000 0 100 100100000 100010010 7; x t, y t, z 0. 00 ; x t, y 0, z t.0b. Not a linear combination. If ax by cz v then comparing entries gives equations 2a b c 4, a c 3 and a b 2c 4. Now carry the coefficient matrix to reduced form: 2 1141 01 031 01 0 1 1 0 1 21 40001Hence there is no solution.d. Here, if aa by cz v then comparing entries gives equations 2a b c 3, a c 0 and a b 2c 3. Carrying the coefficient matrix to reduced form gives 1 01 02 11 31 01 0 0 1 1 3 1 2130000so the general solution is a t, b 3 t and c t. Taking t 1 gives the linear combination v a 2y z.4.b. We must determine if x, y and z exist such that y xa1 ya2 za3 . Equating entries here givesequations x 3y z 1, 3x y z 9, 2y z 2 and x z 6. Carrying the coefficientmatrix to reduced form gives" 1 3 1 1 # " 1 0 0 2 #301120111926000 100 140010so the unique solution is x 2, y 1 and z 4. Hence y 2a1 a2 4a3 .5.b. Carry the augmented matrix to reduced form: 1 1 12 2 2 123101113000 100200010210320000 Hence the general solution is x1 2r 2s 3t, x2 r, x3 s 2t, x4 s and x5 t. In Tmatrix form, the general solution x x1 x2 x3 x4 x5takes the form x 2r 2s 3tr s 2tst r 21000 s 20 110 30 20100100000 t Hence x is a linear combination of the basic solutions.d. Carry the augmented matrix to reduced form:" 1 1 2 2 2 0 #21 22 1 4 428 4410111000 "100001000 2001 300#

8Systems of Linear EquationsHence the general solution x x1 x2 x3 x4 x5 t 0 x 2s 3tst02100 s Tis 13010 t Hence x is a linear combination of the basic solutions. 6.x y 12 has nontrivial solutions with fewer variables than equations.b. The system 2x 2y x y 17.b. There are n r 6 1 5 parameters by Theorem 1.2.2.d. The row-echelon form has four rows and, as it has a row of zeros, has at most 3 leading1’s. Hence rank A r 1, 2 or 3 (r 6 0, because A has nonzero entries). Thus there aren r 6 r 5, 4 or 3 parameters.9.b. Insisting that the graph of ax by cz d 0 (the plane) contains the three points leads tothree linear equations in the four variables a, b, c and d. There is a nontrivial solution byTheorem 1.3.1.11. Since the system is consistent there are n r parameters by Theorem 1.2.2. The system has nontrivialsolutions if and only if there is at least one parameter, that is if and only if n r.1.4 An Application to Network Flows1.b. There are five flow equations, one for each junction: 11000 101001000010000010 100100001000001101 01 0 1 1 1 10 001011 10000f1 f2f1 f3 f5f2 f4 f7 f3 f4 f6f5 f6 f7 1 1 0 0 0 0 0 25 0 00 25 100100 5001 6010 751 1 401 001 00 1 010 101 1 10010010 1011 1000 502535 7540 8560 75 400 0000 11001000010 100 01 00 10 00 00110100111 100001100111If we use f4 , f6 , and f7 as parameters, the solution isf1 85 f4 f700011010 111 1 1 125607540 8560 354040 2550607540

1.5. An Application to Electrical Networks9f2 60 f4 f7f3 75 f4 f6f5 40 f6 f72.b. The solution to (a) gives f1 55 f4 , f2 20 f4 f5 , f3 15 f5 . Closing canal BC meansf3 0, so f5 15. Hence f2 35 f4 , so f2 30 means f4 5. Similarly f1 55 f4 sof1 30 implies f4 25. Hence the range on f4 is 25 f4 30.3.b. The road CD.1.5 An Application to Electrical Networks2. The junction and circuit rules give: 150 11010100Hence1050100510 1 25Left junctionI1 I2 I3Right junction I1 I2 I3Top circuit5I1 10I2Lower circuit10I2 5I3 01 111 11 01 0 15 5 5 0 3 1 1 00105 10021 20 " 1 0 0 1 #1 0 1 15 1 1 0 1 2 14 0 1 0 35 .40I1 15 , I2 3501and I3 5001 0 0 5 10 1121 214545.4. The equations are:Lower left junctionTop junctionMiddle junctionLower right junctionI1 I5 I6 0I2 I4 I6 0I2 I3 I5 0I1 I3 I4 0Observe that the last of these follows from the others (so may be omitted). 100000 011000100000000 11000 10101000 001 0 10 110 000 020 0 1 1 101 1010 100100 1 101 1 11 1 1 121000101020000112Left circuit10I5 10I6 10Right circuit 10I3 10I4 10Lower circuit10I3 10I5 20 1000001000000100000100000010 110010000 1 1 0 101 0 1 1 1 0 01 1 1 100 1010 2 0 1 100 20 2 0102 01 11 50311 2 1 20 12

10Systems of Linear Equations 100000010000001000000100 1 21 213 1000002010000100100000010000001000000112323212 100 1 110 22 215 100000010000001000000100 2 21 3 14000010101012 . Hence I1 2, I2 1, I3 12 , I4 23 , I5 32 , I6 21 . 1.6 An Application to Chemical Reactions2. Suppose xNH3 yCuO zN2 wCu vH2 O where x, y, z, w and v are positive integers. Equatingthe number of each type of atom on each side givesN : x 2zH : 3x 2vCu : y wO:y vTaking v t these give y t, w t, x 23 t and z 12 x 13 t. The smallest value of t such that thereare all integers is t 3, so x 2, y 3, z 1 and v 3. Hence the balanced reaction is2NH3 3CuO N2 3Cu 3H2 O4. 15Pb(N3)2 44Cr(MnO4 )2 22Cr2O3 88MnO2 5Pb3 O4 90NOSupplementary Exercises: Chapter 11.2.b. No. If the corresponding planes are parallel and distinct, there is no solution. Otherwise theyeither coincide or have a whole common line of solutions. 6616" 1 4 1 1 2 # " 1#1 01010104 112 0 1 4 1 1 4 1 1b. 31 62 13 20 51 00 10 0 0 010 100 100 . 104 1 1114 5230 4101100000110 (16 6s 6t)3.Hence x3 s, x4 t are parameters, and the equations give x1 110 (1 4s t). 113a113a1 1 3 ab. a 1 5 4 0 1 a 5 3a 4 a2 0 1 a 5 3a 4 a22 .1 a 4 a0 a 110003(2 a) 4 a 1 1 3 11 1 3 0If a 1 the matrix is 0 0 2 3 0 0 1 1 , so there is no solution.00330001and x2

1.6. An Application to Chemical ReactionsIf a 2 the matrix is 1001 102003 10 100010200210If a 6 1 and a 6 2 there is a unique solution.# " " 1 1 3aa1132a 40 1 a5 3a4 a2 0 1 3a 5a 1a 1203(2 a)0Hence x 4 a8 5a,3(a 1)y 0 a 2,3(a 1)0z a 231 11, so x 2 2t, y t, z t.10012a 13a 5a 1001 a 4a 1a2 4a 1a 23#a 23 . 100010000 5a 83(a 1) a 23(a 1)a 23 .4. If R1 and R2 denote the two rows, then the following indicate how they can be interchanged usingrow operations of the other two types: R1R1 R2R1 R2R2R2 R2R2 R1 R1R1Note that only one row operation of Type II was used — a multiplication by 1.6. Substitute x 3, y 1 and z 2 into the given equations. The result is3 a 2c 03b c 6 13a 2 2b 5 2c 33b c 93a 2b 7athat isThissystem of linear equations for a, b and c has unique solution: 1 0 231 0 231 01 0 2 30 3 1 7 0 3 1 7 0 1 7 9 0 13207026Hence a 1, b 2, c 1. 111 5112 1 1 18. 0 3 322005 21 3505 915 2 100 26110110530 00 2 72010001001039 20230 . 10001000112 1 .Hence the solution is x 2, y 3 t, z t. Taking t 3 i gives x 2, y i, z 3 i, as required.If the real system has a unique solution, the solution is real because all the calculations in thegaussian algorithm yield real numbers (all entries in the augmented matrix are real).

2. Matrix Algebra2.1 Matrix Addition, Scalar Multiplication, andTransposition1.2.b. Equating entries gives four linear equations: a b 2, b c 2, c d 6, d a 2. Thesolution is a 2 t, b 4 t, c 6 t, d t.d. Equating coefficients gives: a b, b c, c d, d a. The solution is a b c d t, tarbitrary.hih ihi hi h i hi hi hi79 30 7 14b. 3 13 5 62 7 11 39 30 10 7 3 10 7 20d. [ 3 1 2 ] 2 [ 9 3 4 ] [ 3 11 6 ] [ [ 3 18 3 1 6 11 2 8 6 ] [ 12 4 0 12 T01 210 4 1 0 4f. 2h. 33.h2 1b. 5C 5410h32 2 10 420iTih12 h 131510i 122]] [1868] [ 63 11]0 3 503 1ih21 10i 2h12 13i h63 30i h24 26i d. B D is not defined as B is 2 3 while D is 3 2.hiT hiT hi 31 15052f. (A C)T 20 2 1 02 10 14.4 1 1 6ih. A D is not defined as A is 2 2 while D is 3 2.h ih ih ih ib. Given 3A 21 5A 2 30 , subtract 3A from both sides to get 21 2A 2 30 . Nowh ih ih i h i32add 2 0 to both sides: 2A 1 2 30 81 . Finally, multiply both sides by 12 :h i h i4A 12 81 1 .25.hb. Given 2A B 5(A 2B), add B to both sides to get2A 5(A 2B) B 5A 10B B 5A 11BNow subtract 5A from both sides: 3A 11B. Multiply by 13 to get A 113 B.13

14Matrix Algebra4X 3Y A, subtract the first from the second to get X Y B A. Now subtract 35X 4Y Btimes this equation from the first equation: X A 3(B A) 4A 3B. Then X Y B Agives Y (B A) X (B A) (4A 3B) 4B 5A.Note that this also follows from the Gaussian Algorithm (with matrix constants):hi hi hi4 3 A5 4 B1 1 B A 5 4 B4 3 A4 3Ahi hi1 0 4A 3B11B A 0 1 5A 4B 0 1 4B 5A6.b. Given7. b. Given 2X 5Y 1 2 let Y T where T is an arbitrary 1 2 matrix. Then 2X 5T 1 2 so X 25 T 12 1 2 , Y T . If T s t , this gives X 52 s 12 25 t 1 , Y s t , where s and t are arbitrary.8.9.b. 5[3(A B 2C) 2(3C B) A] 2[3(3A B C) 2(B 2A) 2C] 5[3A 3B 6C 6C 2B A] 2[9A 3B 3C 2B 4A 2C] 5[2A B] 2[5A B C] 10A 5B 10A 2B 2C 20A 7B 2Chib. Write A ac db . We want p, q, r and s such thathacbdi ph1001i qh1010i rh1100i sh0110Equating com

Linear Algebra with Applications Open Edition Base Text Revision History Current Revision: Version2019 — RevisionA 2019 A New Section on Singular Value Decomposition(8.6) is included. NewExample2.3.2andTheorem2.2.4. Please notethatthis will impactthenumberingofs