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601-hw-weeks3-4

601-hw-weeks3-4 - AN2 Answers to Selected Odd-Numbered...

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Unformatted text preview: AN2 Answers to Selected Odd-Numbered Exercises 1 l 2 6 B: 3 4—1 —1 l l 2 l l l l l l l 29.A=23I.B=23l2 1—1 3 l—l 3 2 31.X|+2Xz— X3: —Xz+3X3=l 5x2—2x3=6 33.X1+ x2: 9 —-2x2= —2 -2x2=—21 35.x. +2xz— x3+x4=1 x2+ X3—x4=3 312+6X3 =1 Exercises 1.2, p. 26 l. a) The matrix is in echelon form. b) The operation R. — 21?; yields reduCCd I 0 echelon form . 0 l 3. a) The operations R2 — 2R., (l/2)R., R2 4 4R., (1/5)R2 yield echelon form 1 3/2 1/2 0 1 2/5 ' 5. a) The operations R. 9 R2. (l/2)R., (l/2)R2 yield echelon form 1 0 1/2 2 0 0 1 3/2 ' 7. a) The matrix is in echelon form. b) The operations R. — 2R3, R2 — 4R3. R. — 3R2 yield reduced echelon form 005 0—2 1 01 001] 9. a) The operation (1/2) R2 yields echelon form 1 2—1 —2 0 1—1—3/2 0 o o I 11.x. =0,x2=0 13. x. = —2 + 5x;., x2 = l — 3x3. x3 arbitrary 15. The system is inconsistent. 17. x. = x3 = x4 = 0; x2 arbitrary 19. The system is inconsistent. 21. x. = —l — (1/2)x2 + (1/2)x4, x3 = l — X4, x2 and x4 arbitrary, X5 = O 23. Inconsistent 25. x. = 2 — x2, x2 arbitrary 27. x. = 2 — x2 + x3, x2 and x3 arbitrary 29. x. = 3 - 2x3, x2 = —-2 + 3X3, 153 arbitrary 31. x. = 3 — (7X4 — 16I5)/2, x2 = (X4 + 2X5)/2, x3 = —2 + (5x. — 12x5)/2, x4 and x5 arbitrary 33. Inconsistent 35. Inconsistent 37. All values of a excepta = 8 39.a=3ora=—3 41. a = n/3 ora = SIT/3; ﬂ = 7r/6 orﬂ = 571/6 45.lxx lxx lxx [01x]’[001]’[000]’ 01x 01x 001 [001]'[000]’[000]’ 000 [000] 47. The operations R2 —2R., R. +2R2, —R2 transform B to l. The operations R2 — 3R. , R. + R2, (—1/2)R2 reduce C to I , so the operations —2R2, R. — R2, R2 +3R. trans- form I to C. Thus the operations R2 — 2R., R. + 2R2, —R2, —2R2. R. — R2. R2 + 3R. transform B to C. 49. N = 135 51. The amounts were \$39. \$21, and \$12. 53. Let A denote the number of adults, S the number of stu- dents, and C the number of children. Possible solutions are: A = 5k, S = 67~llk, C =12+6k, wherek = 0, 1, . . . , 6. 55. n(n +1)/2 57. n(n + l)(2n + 1)(3n2 +3n — 1)/30 S9. 61. n=4, m=6 n=5. m=5 Exercises 1.7, p. 78 1. 9:45"? ll. 13. 15. 17. 21. 23. 25. 27. 31. 35. 39. 41. 45. 47. Linearly independent Linearly dependent, v5 = 3v, Linearly dependent, v3 = 2v, Linearly dependent, u4 = 4115 Linearly independent Linearly dependent, u4 = 4u5 16 12 4 Linearly dependent, u4 = gun + ?u. - Eu; Those in Exercises 5. 6, l3, and 14 Singular; x. = —2x2 19. Singular; x. = —2xz Singular; x. = x2 = 0, x3 arbitrary Nonsingular Singular; x2 = x3 = 0. x] arbitrary Nonsingular 29. a = 6 b(a—2)=4 33.c—ab=0 V3 = A2 37. V2 = (C| + U3 =(-8F1— 2F2 + b=—11V1+7V2 0V|+0V2 b = —3V1+2V2 a) Any value a b) Any value a Exercises 1.8, p. 90 l. 3. 5. 7. PO) = (-1/2)I2 + (9/2)! -1 p(t) = 2: + 3 p(t) = 213 — 212 + 3! +1 y = 2122‘ + e” 9. y = 3wr +4e" + ch 11. 3h f(t)dt 7*" 7”“) + f(2h)] 0 3h 13. f (t) d: 0 3!: x ?mm + 3m) + 3mm + mm] It I 15. f0 f(t)dt a» §l[—f(—h) + 3 f(0)] 17. f’(0) av [—f(0) + f (h)]/ h 19. f’(0) z [—3f(0) + 4f(h) - f(2h)]/(2h) 21. NO) 5: m—h) — 2f(0) + fun/h2 Answers to Selected Odd-Numbered Exercises AN5 .0 27. p(t) =13 +2:2 +3: +2 29.p(l)=t3+!2+4t+3 1 5. ' ~— 3 f“) 12h x [f(a-2h)—8f(a—h)+8f(a+h)—f(a+2h)] Exercises 1.9, p. 102 5. I] = -3, X2 = 7. I; = 14, 9. If B = (by) is a (3 x 3) matrix such that AB = I, then 0b” +Ob2. +0b31 = 1. Since this is impossible, no such matrix exists. 13. 3 —l —2 1 15. —1/3 2/3 17. I 0 0 0 1 X2 = —20,I3 = 8 2/3 —1/3 -2 1 5 —4 l9. 1 —2 0 3—3 —1 —6 7 2 21. —1/2—2/3 —I/6 7/6 1 1/3 1/3 —4/3 0 —1/3 —1/3 1/3 —1/2 1 1/21/2 23.A-'=(1/10)[ 3 2] —2 2 27.A=2and}t= —2 31. .1" =18, 25. A has no inverse 29. XI =6, X2=—8 33. x, = 5/2, x; = 5/2 35. Q" =C-'A-' =[ _3 l ] x2=13 3 5 37. Q_l =(A'I)T=[ 3 0] l 2 39. Q" = (A-I)T(c-I)T =[ f 2’ :| 1 5 41. Q-1 = BC" = —1 4 3/2 1/2 ] 43. Q" = (1/2M-I = [ 0 1 480 Solutions to Selected Exercises 1 l l/3 0 3 3 304—16 2 3 = Amxn = meanxn' 2 1 2/3 —1/./2 3.4.17 R)? = QTb s i = R" lQT!) = (5/9, 0). 3.4.18 Q has the same column spaCe as A, so P = Q(QTQ)’ lQT = QQT. 2 02“ y(x) sin x dx. 3.4.20 1; ./(e — 1 )/2; 3.4.22 22 ,0. f0 sinzx dx 3.4.24 x3 is already orthogonal to the even functions 1 and x2 —- %. To subtract oiT its com- ponent in the direction of the odd function x, we compute (x3, x) = [1, 2:31: dx = ‘ §and (x, x) = it, x2 dx = %; the next Legendre polynomial is x3 — ((x’, x)/(x, x))x = x3 — 35x. 3.4.27 I/ﬁ. —1/ﬁ.0.o). (1/f6. 1/J3.2/J6.o), (— 1W3. —1/2J§. 1/2«/§. — W3). 400016000 0 0 0 4 0 l6 0 0 3.5.1 0 0 4 0 ’ 0 0 16 0 ' 0 4 0 0 0 0 0 16 3.5.2 1, —1/2 + (ﬁ/zy, —1/2 —(./3/2)1. 3.5.5 x = (2k + l)n, 9 = 2k7r + 11/2, k is integer. 3.5.7 (1,0,1,0). 3.5.8 (0,101). 35.10 .Vo =yb+y3.y1 = 1221-118; .Vo =14. +113, y. =y'1 + iy’l’. yz =yb —y3. ya =y'. —1'y’1’- 1 1 2 2 0 1 o 0 3.5.1] I —> 0 -+ 0 —> 2 . 0 0 0 0 3513 “0 = (/0 +f1+f2 +f3)/4: C1=lfo " ifl ‘fz + ifsi/‘L 0: = (/0 —fl +f2 ‘13)“. _C3 Tlfo :difl ‘fz - Val/‘kfo = Oafz = Oafs = ‘fi =’ “o = 0,02 = 0’53 = -‘31=> C 13330 0 . 3.6.1 (8)7 (b)2 (c)8 (d) 13. “11 “12 “13 “14 3.6.3 V+Wcontains the 4 by4 matrices ““ a“ “23 “24 ,dim(V+W)=l3, 0 “32 “33 “34 0 0 “43 “44 a“ an 0 0 0 “22 “23 V n Wcontains the matrices , dim(V n W) = 7, 0 “23 “34 0 0 0 a44 dim(V+ W)+dim(Vn W)=20=dim V+dim W. . 3.6.6 The line through (1,0, 0) (it need not be perpendicular to V). 3.6.7 x=vl+wl=v2+w2, 01,026V,wbwzewav,-—vz=wz—w,eVnW= {0}=vl=vz,wl=w2. Solutions to Selected Exercises 481, 3.6.9 A = ‘1’ 8], B = [3 (1,], nullspace of AB = [3 ‘1’] does not contain [‘2], column space of AB is not contained in column space of B. 3.6.11 rank(AB) s rank(A), rank(A) s n, n < m => rank(AB) < m = AB is singular. 3.6.12 rank(A + B) = dim(ﬁM) + @(B)) S dim 99M) + dim 92(8) = rank(A) + rank(B). 3.6.13 .A/(B) = .A’(A'1AB) 2 JV(AB) 2 ./V(B) => .A/(AB) = = .A’(B); the row space is the orthogonal complement, and the rank is its dimension—therefore these are the same for B and AB. 1 0 1 __1 0 l 0 3.6.15 A: 0 1 = 0 [l —-1 0]+ l [0 I —1]. 0 1 —1 l l 1 1 3.6.16 dim(U n V) = dim(U) + dim(V) — dim(U u V) =: dim(U n V) 2 6 + 6 —- 8 = 4 > 0, dim(U n V n W) = dim(U n V) + dim(W) ~— dim((U n V) u W) => dim(Un Vn W)24+6—-8=2>0. 3.6.18 ye Vn W¢>y = xiv, + - - - + xkv,‘ and also y = —x,..,lwl — ' - - -—x,‘+,wl (*)¢> x1v1+~ - + kaW, = 0 ¢> x = (x1, . . . , ka) e .xV(D). Because of the uniqueness of the expressions (*) y yields only one x in .A" (D). w§b1 + ' ' ' + wﬁbm w: + - - - + w; ' 3‘6‘22 j-l'V = (211—, z (211-, 2%: b _ = (“-2113 .211! —.2él- a ‘- Af’y) = 0. 3-6-23 (a) E(e)=%(-2)+ %(-1)+ H5) = 0.1:"(ezl=%(-2)2 +;‘i(-1)2 + M52) = J‘21 (b) “’1 = Jig, “’2 = 2' 3.6.24 A = B + C, with B containing those p rows and zero elsewhere, C = A — B. rank(A) s rank(B) + rank(C) S p + q; 5 by 5. 3.6.20 fw = 3.6.21 11; 5; the line through (1, —4). CHAPTER 4 4.2.1 2" dct(A); (—1)" dct(A); (det(A))2. 4.2.3 20; 5. 4.2.6 (a) 0 (b) 16 (c) 16 ((1) 1/16 (e) 16. 4.2.7 Multiply the zero row by 2. That multiplies det A by 2, but A is unchanged => det A = 0. 4.2.8 (1 — ml)(ad — bc). 1 a a2 1 a a2 l a (12 4.2.10 1 b b2=0 b—a bZ—az=0 b—a bz—az = l c c2 0 c—a «:2—a2 0 0 (c—a)(c—b) (b — a)(c — a)(c — b). 4.2.12 (a) False; det[‘f }] aé 2 det [f i] (b) False; det[‘,’ }] = — 1, its pivots are l, l, but there is a row exchange (c) True; dct(AB) = dct(A) dct(B) == 0. 4.2.13 Adding every column of A to the ﬁrst column makes it a zero column, so det A = 0. If every row of A adds to 1, every row of A —1 adds to 0 => dct(A —- 1) =0; 1 1 A =[2 i],det(A—1)=O,butdetA=0=,é 1. l 2 2 ...
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