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Instructor’s Manualto accompanyFundamental MethodsofMathematical EconomicsFourth EditionAlpha C. ChiangUniversity of ConnecticutKevin WainwrightBritish Columbia Institute of Technology

Title of Supplement to accompanyFUNDAMENTAL METHODS OF MATHEMATICAL ECONOMICSAlpha C. Chiang, Kevin WainwrightPublished by McGraw-Hill, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,New York, NY 10020. Copyright 2005, 1984, 1974, 1967 by The McGraw-Hill Companies, Inc.All rights reserved.The contents, or parts thereof, may be reproduced in print form solely for classroom use withFUNDAMENTAL METHODS OF MATHEMATICAL ECONOMICSprovided such reproductions bear copyright notice, but may not be reproduced in any other form or for anyother purpose without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limitedto, in any network or other electronic storage or transmission, or broadcast for distance learning.ISBN 0-07-286591-1 (CD-ROM)www.mhhe.com

ContentsCONTENTS1CHAPTER 26Exercise 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6Exercise 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6Exercise 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7CHAPTER 39Exercise 3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9Exercise 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9Exercise 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .10Exercise 3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11CHAPTER 413Exercise 4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13Exercise 4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14Exercise 4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15Exercise 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17Exercise 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19Exercise 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20Exercise 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21CHAPTER 522Exercise 5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22Exercise 5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23Exercise 5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .24Exercise 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25Exercise 5.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26Exercise 5.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27Exercise 5.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29CHAPTER 6321

Exercise 6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32Exercise 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .32Exercise 6.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Exercise 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Exercise 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34CHAPTER 735Exercise 7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35Exercise 7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35Exercise 7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37Exercise 7.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37Exercise 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .38Exercise 7.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39CHAPTER 840Exercise 8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40Exercise 8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41Exercise 8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43Exercise 8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .44Exercise 8.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45Exercise 8.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .47CHAPTER 951Exercise 9.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51Exercise 9.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51Exercise 9.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52Exercise 9.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .54Exercise 9.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .55CHAPTER 1056Exercise 10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56Exercise 10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56Exercise 10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57Exercise 10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .582

Exercise 10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59Exercise 10.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .60Exercise 10.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61CHAPTER 1163Exercise 11.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .63Exercise 11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .64Exercise 11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66Exercise 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .69Exercise 11.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71Exercise 11.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .73CHAPTER 1276Exercise 12.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .76Exercise 12.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .77Exercise 12.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .78Exercise 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .81Exercise 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83Exercise 12.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .85CHAPTER 1387Exercise 13.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .87Exercise 13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .88Exercise 13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91CHAPTER 1492Exercise 14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .92Exercise 14.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .93Exercise 14.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95Exercise 14.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .96Exercise 14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .97CHAPTER 1598Exercise 15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .398

Exercise 15.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99Exercise 15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100Exercise 15.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Exercise 15.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Exercise 15.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Exercise 15.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104CHAPTER 16106Exercise 16.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106Exercise 16.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107Exercise 16.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109Exercise 16.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110Exercise 16.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112Exercise 16.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114Exercise 16.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115CHAPTER 17117Exercise 17.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117Exercise 17.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Exercise 17.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118Exercise 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120Exercise 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121CHAPTER 18123Exercise 18.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123Exercise 18.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124Exercise 18.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125Exercise 18.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126CHAPTER 19129Exercise 19.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129Exercise 19.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131Exercise 19.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133Exercise 19.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1354

Exercise 19.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138CHAPTER 20141Exercise 20.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s ManualCHAPTER 2Exercise 2.31.(a) {x x 34}(b) {x 8 x 65}2. True statements: (a), (d), (f), (g), and (h)3.(a) {2,4,6,7}(b) {2,4,6}(c) {2,6}(d) {2}(e) {2}(f) {2,4,6}4. All are valid.5. First part: A (B C) {4, 5, 6} {3, 6} {3, 4, 5, 6} ; and (A B) (A C) {3, 4, 5, 6, 7} {2, 3, 4, 5, 6} {3, 4, 5, 6} too.Second part: A (B C) {4, 5, 6} {2, 3, 4, 6, 7} {4, 6} ; and (A B) (A C) {4, 6} {6} {4, 6} too.6. N/A7. , {5}, {6}, {7}, {5, 6}, {5, 7}, {6, 7}, {5, 6, 7}8. There are 24 16 subsets: , {a}, {b}, {c}, {d}, {a,b}, {a,c}, {a,d}, {b,c}, {b,d}, {c,d},{a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}, and {a,b,c,d}.9. The complement of U is Ũ {x x / U }. Here the notation of ”not in U ” is expressed via the / symbol which relates an element (x) to a set (U ). In contrast, when we say ” is a subsetof U,” the notion of ”in U” is expressed via the symbol which relates a subset( ) to a set(U ). Hence, we have two different contexts, and there exists no paradox at all.Exercise 2.41.(a) {(3,a), (3,b), (6,a), (6,b) (9,a), (9,b)}(b) {(a,m), (a,n), (b,m), (b,n)}(c) { (m,3), (m,6), (m,9), (n,3), (n,6), (n,9)}2. {(3,a,m), (3,a,n), (3,b,m), (3,b,n), (6,a,m), (6,a,n), (6,b,m), (6,b,n), (9,a,m), (9,a,n), (9,b,m),(9,b,n),}6

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual3. No. When S1 S2 .4. Only (d) represents a function.5. Range {y 8 y 32}6. The range is the set of all nonpositive numbers.7. (a) No.(b) Yes.8. For each level of output, we should discard all the inefficient cost figures, and take the lowestcost figure as the total cost for that output level. This would establish the uniqueness asrequired by the definition of a function.Exercise 2.51. N/a2. Eqs. (a) and (b) differ in the sign of the coefficient of x; a positive (negative) sign means anupward (downward) slope.Eqs. (a) and (c) differ in the constant terms; a larger constant means a higher vertical intercept.3. A negative coefficient (say, -1) for the x2 term is associated with a hill. as the value of x issteadily increased or reduced, the x2 term will exert a more dominant influence in determiningthe value of y. Being negative, this term serves to pull down the y values at the two extremeends of the curve.4. If negative values can occur there will appear in quadrant III a curve which is the mirror imageof the one in quadrant I.5.(a) x196.(a) x6(b) xa b c(c) (xyz)3(b) x1/67. By Rules VI and V, we can successively write xm/n (xm )1/n we also have xm/n (x1/n )m ( n x)m nxm ; by the same two rules,8. Rule VI:mn(xm )n xm xm{z . xm} x x {z . x} xn term smn term s7

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsRule VII:xm y m x x . x y y . . . y{z} {z}m term sm term s (xy) (xy) . . . (xy) (xy)m{z} m term s8Instructor’s Manual

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s ManualCHAPTER 3Exercise 3.231. (a) By substitution, we get 21 3P 4 8P or 11P 25. Thus P 2 11. Substituting2P into the second equation or the third equation, we find Q 14 11.(b) With a 21, b 3, c 4, d 8, the formula yieldsP 25113 2 11Q 156112 14 112.(a)P 619 6 79Q 2769 30 23P 367 5 17Q 1387 19 57(b)3. N/A4. If b d 0 then P and Q in (3.4) and (3.5) would involve division by zero, which is undefined.5. If b d 0 then d b and the demand and supply curves would have the same slope(though different vertical intercepts). The two curves would be parallel, with no equilibriumintersection point in Fig. 3.1Exercise 3.31.(a) x 1 5;x 2 3(b) x 1 4;x 2 22.(a) x 1 5;x 2 3(b) x 1 4;x 2 23.(a) (x 6)(x 1)(x 3) 0, or x3 8x2 9x 18 0(b) (x 1)(x 2)(x 3)(x 5) 0, or x4 11x3 41x2 61x 30 04. By Theorem III, we find:(a) Yes.(b) No.(c) Yes.9

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual5.(a) By Theorem I, any integer root must be a divisor of 6; thus there are six candidates: 1, 2, and 3. Among these, 1, 12 , and 14(b) By Theorem II, any rational root r/s must have r being a divisor of 1 and s being adivisor of 8. The r set is {1, 1}, and the s set is {1, 1, 2, 2, 4, 4, 8, 8}; these giveus eight root candidates: 1, 12 , 14 , and 18 . Among these, 1, 2, and 3 satisfy theequation, and they constitute the three roots.(c) To get rid of the fractional coefficients, we multiply every term by 8. The resultingequation is the same as the one in (b) above.(d) To get rid of the fractional coefficients, we multiply every term by 4 to obtain4x4 24x3 31x2 6x 8 0By Theorem II, any rational root r/s must have r being a divisor of 8 and s being adivisor of 4. The r set is { 1, 2, 4, 8}, and the s set is { 1, 2, 4}; these give usthe root candidates 1, 12 , 14 , 2, 4, 8. Among these, 12 , 12 , 2, and 4 constitute thefour roots.6.(a) The model reduces to P 2 6P 7 0. By the quadratic formula, we have P1 1 andP2 7, but only the first root is acceptable. Substituting that root into the second orthe third equation, we find Q 2.(b) The model reduces to 2P 2 10 0 or P 2 5 with the two roots P1 Only the first root is admissible, and it yields Q 3. 5 and P2 5.7. Equation (3.7) is the equilibrium stated in the form of ”the excess supply be zero.”Exercise 3.41. N/A10

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual2.P1 (a2 b2 )(α0 β 0 ) (a0 b0 )(α2 β 2 )(a1 b1 )(α2 β 2 ) (a2 b2 )(α1 β 1 )P2 (a0 b0 )(α1 β 1 ) (a1 b1 )(α0 β 0 )(a1 b1 )(α2 β 2 ) (a2 b2 )(α1 β 1 )3. Since we havec0 18 2 20γ 0 12 2 14it follows thatc1 3 4 7c2 1γ1 1γ 2 2 3 5P1 14 10035 1 57176 3 17andP2 20 9835 1 59178 3 17Substitution into the given demand or supply function yieldsQ 1 194177 11 17andQ 2 143177 8 17Exercise 3.51.(a) Three variables are endogenous: Y, C, and T.(b) By substituting the third equation into the second and then the second into the first, weobtainY a bd b(1 t)Y I0 G0or[1 b(1 t)]Y a bd I0 G0ThusY a bd I0 G01 b(1 t)Then it follows that the equilibrium values of the other two endogenous variables areT d tY d(1 b) t(a I0 G0 )1 b(1 t)andC Y I0 G0 11a bd b(1 t)(I0 G0 )1 b(1 t)

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual2.(a) The endogenous variables are Y, C, and G.(b) g G/Y proportion of national income spent as government expenditure.(c) Substituting the last two equations into the first, we getY a b(Y T0 ) I0 gYThusY a bT0 I01 b g(d) The restriction b g 6 1 is needed to avoid division by zero.3. Upon substitution, the first equation can be reduced to the formY 6Y 1/2 55 0orw2 6w 55 0(where w Y 1/2 )The latter is a quadratic equation, with roots 1w1 , w2 6 (36 220)1/2 11, 52From the first root, we can getY w1 2 121andC 25 6(11) 91On the other hand, the second root is inadmissible because it leads to a negative value for C:C 25 6( 5) 512

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s ManualCHAPTER 4Exercise 4.11.Qd Qs Coeffi cient M atrix: 1 10 10b 01 d 0QdQs bP a dP cVector of Constants: 0 a c2.Qd1 Qs1 0Qd1Qs1Qd2Qs2 Coeffi cient m atrix:1 1 000010 00 a1 a201 00 b1 b200 1 10000 10 α1 α200 01 β 2 β 1 a2 P2 a0 b1 P1 b2 P2 b0 Qs2Qd2 a1 P1 0 α1 P1 α2 P2 α0 β 1 P1 β 2 P2 β0Variablevector: Qd1Qs1Qd2Qs2P1P2Constantvector: 3. No, because the equation system is nonlinear4.Y C bY CThe coefficient matrix and constant vector are 1 1 b113 I0 G0 a I0 G0a 0 a0 b0 0 α0 β0

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual5. First expand the multiplicative expression (b(Y T ) into the additive expression bY bT sothat bY and bT can be placed in separate columns. Then we can write the system asY C I0 G0 C a bY bT tY T d Exercise 4.21.2. (a) 7 39 7 (b) 140 828 64 (c) 21 31827 (d) 162224 6 (a) Yes AB 6 0 . No, not conformable. 13 8 20 1614 4 6 CB (b) Both are defined, but BC 21 2469 30 15 12100 35 610 1 0 0 314 3. Yes. BA 3 15 28 0101 2 10510 24620 0 105 105 10Thus we happen to have AB BA in this particular case. 0 2 hi4933x 5y (c) (d) 7a c 2b 4c4. (a) 36 20 (b) 4 34x 2y 7z(1 2)16 3(2 2)(2 1)(3 2)5. Yes. Yes.Yes. Yes.6.(a) x2 x3 x4 x5(b) a5 a6 x6 a7 x7 a8 x8(c) b(x1 x2 x3 x4 )(d) a1 x0 a2 x1 · · · an xn 1 a1 a2 x a3 x2 · · · an xn 114

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual(e) x2 (x 1)2 (x 2)2 (x 3)27.(a)3Pi 18.(a)(b)µixi (xi 1)nPi 1xi¶(b)4Pai (xi 1 i)(c)i 2nPi 11xi xn 1 x0 x1 · · · xn xn 1 nXj 1abj yj(d)nPi 0n 1P1xixii 1 ab1 y1 ab2 y2 · · · abn yn a(b1 y1 b2 y2 · · · bn yn ) anXbj yjj 1(c)nX(xj yj ) (x1 y1 ) (x2 y2 ) · · · (xn yn )j 1 (x1 x2 · · · xn ) (y1 y2 · · · yn )nnXX xj yjj 1j 1Exercise 4.31. 5 15 5 5 hi (a) uv 0 1 3 1 1 3 1 1 9 3 33 35 25 405 hi (b) uw0 1 5 7 8 1 7 5 8 21 15 243 x2x1x1 x2 x1 x3 h i 1 0(c) xx x2 x1 x2 x3 x2 x1 x22x2 x3 x3x3 x1 x3 x2 x23 5hi 0(d) v u 3 1 1 1 [15 1 3] [44] 44 315

Chiang/Wainwright: Fundamental Methods of Mathematical Economics3 1 [15 1 3] 13 1 x1 hi (f) w0 x 7 5 8 x2 [7x1 5x2 8x3 ] 7x1 5x2 8x3 x3 5hi (g) u0 u 5 1 3 1 [25 1 9] [35] 35 3 x1 hi 3 P (h) x0 x x1 x2 x3 x2 x21 x22 x23 x2i i 1x3(e) u0 v 2. Instructor’s Manualh5 1 3i (a) All are defined except w0 x and x0 y 0 . hixxyxy11112 y1 y2 (b) xy 0 x2x2 y1 x2 y2 hi y1 y12 y22xy 0 y1 y2 y2 2hizzzz112 z1 z2 1 zz 0 z2z2 z1 z22 hi2y16yy3y111 3 2 16 1 yw0 y23y2 2y2 16y2x · y x1 y1 x2 y23.(a)nPPi Qii 1(b) Let P and Q be the column vectors or prices and quantities, respectively. Then the totalrevenue is P · Q or P 0 Q or Q0 P .16

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual4.(a) w10 w2 11 (acute angle, Fig. 4.2c)(b) w10 w2 11 (obtuse angle, Fig. 4.2d)(c) w10 w2 13 (obtuse angle, Fig. 4.2b)(d) w10 w2 0 (right angle, Fig. 4.3)(e) w10 w2 5 (acute angle, Fig. 4.3) 055. (a) 2v (b) u v 64 5 (e) 2u 3v (d) v u 26. 1011(a) 4e1 7e2(b) 25e1 2e2 e3(c) e1 6e2 9e3(d) 2e1 8e3 (c) u v 5 2 20 (f) 4u 2v 2 7.p (3 0)2 (2 1)2 (8 5)2 27p (b) d (9 2)2 0 (4 4)2 113(a) d 8. When u, v, and w all lie on a single straight line.9. Let the vector v have the elements (a1 , . . . , an ). The point of origin has the elements (0, . . . , 0).Hence:(a)d(0, v) d(v, 0)(b) d(v, 0) (v 0 v)1/2p(a1 0)2 . . . (an 0)2p a21 . . . a2n [See Example 3 in this section](c) d(v, 0) (v · v)1/2Exercise 4.41. (a) (A B) C A (B C) 51711 1717

Chiang/Wainwright: Fundamental Methods of Mathematical Economics (b) (A B) C A (B C) 199 12. No. It should be A B B A 250 68 3. (AB)C A(BC) 75 55(a)k(A B)Instructor’s Manual k[aij bij ] [kaij kbij ] [kaij ] [kbij ] k[aij ] k [bij ] kA kB(b)(g k)A (g k)[aij ] [(g k)aij ] [gaij kaij ] [gaij ] [kaij ] g [aij ] k [aij ] gA kA4.(a)AB (12 3) (14 0) (12 9) (14 2)(20 3) (5 0) 36 136 60 190(20 9) (5 2) (b)AB (4 3) (7 2) (4 8) (7 6) (4 5) (7 7)(9 3) (1 2) (9 8) (1 6) (9 5) (1 7) 26 74 69 29 78 52 (c)AB (7 12) (11 3) (7 4) (11 6) (7 5) (11 1) (2 12) (9 3) (2 4) (9 6) (2 5) (9 1) (10 12) (6 3) (10 4) (6 6) (10 5) (6 1) 117 94 46 51 62 19 C 138 76 5618

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual(d) AB(6 10) (2 11) (5 2) (6 1) (2 3) (5 9)(7 10) (9 11) (4 2) (7 1) (9 3) (4 9) 92 57 177 70(e) 2 3 2 6 2 2 i. AB 4 3 7 34 67 6 6 12 4 2 12 217 2ii. BA [(3 2) (6 4) ( 2 7)] [4]24424 8 145. (A B)(C D) (A B)C (A B)D AC BC AD BD6. No, x0 Ax would then contain cross-product terms a12 x1 x2 and a21 x1 x2 .7. Unweighted sum of squares is used in the well-known method of least squares for fitting anequation to a set of data. Weighted sum of squares can be used, e.g., in comparing weatherconditions of different resort areas by measuring the deviations from an ideal temperature andan ideal humidity.Exercise 4.51. 1 1 x1(a) AI3 (b) I2 A (c) I2 x (d) x0 I2 h5 70 2 45 70 2 4 x2x1 x2 i19

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual2. (a) Ab 9 30 0 0 12 021 12 (b) AIb gives the same result as in (a).hi(c) x0 IA x1 5x1 2x2 7x1 4x2(d) x0 A gives the same result as in (c)3.(a) 5 3(b) 2 6(c) 2 1(d) 2 54. The given diagonal matrix, when multiplied by itself, gives another diagonal matrix with thediagonal elements a211 , a222 , . . . , a2nn . For idempotency, we must have a2ii aii for every i. Henceeach aii must be either 1, or 0. Since each aii can thus have two possible values, and sincethere are altogether n of these aii , we are able to construct a total of 2n idempotent matricesof the diagonal type. Two examples would be In and 0n .Exercise 4.61.2. A0 0 143 B0 (a) (A B)0 A0 B 0 3 0 8 13 1 43 1 6 C0 0 1 9 1 2417 (b) (AC)0 C 0 A0 43 4 6 3. Let D AB. Then (ABC)0 (DC)0 C 0 D0 C 0 (AB)0 C 0 (B 0 A0 ) C 0 B 0 A0 1 0 , thus D and F are inverse of each other, Similarly,4. DF 0 1 1 0 , so E and G are inverses of each other.EG 0 1 5. Let D AB. Then (ABC) 1 (DC) 1 C 1 D 1 C 1 (AB) 1 C 1 (B 1 A 1 ) C 1 B 1 A 120

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s Manual6.(a) A and X 0 X must be square, say n n; X only needs to be n m, where m is not necessarilyequal to n.(b)AA [I X(X 0 X) 1 X 0 ][I X(X 0 X) 1 X 0 ] II IX(X 0 X) 1 X 0 X(X 0 X) 1 X 0 I X(X 0 X) 1 X 0 X(X 0 X) 1 X 0[see Exercise 4.4-6] I X(X 0 X) 1 X 0 X(X 0 X) 1 X 0 XI(X 0 X) 1 X 00 1 I X(X X)X[by (4.8)]0 AThus A satisfies the condition for idempotency.Exercise 4.71. It is suggested that this particular problem could also be solved using a spreadsheet or othermathematical software. The student will be able to observe features of a Markov process morequickly without doing the repetitive calculations. 0.9 0.1 (a) The Markov transition matrix is 0.7 0.3(b)Two periodsThree PeriodsFive PeriodsTen 150(c) As the original Markov transition matrix is raised to successively greater powers theresulting matrix converges toMnn 0.875 0.1250.875 0.125 which is the ”steady state”, giving us 1050 employed and 150 unemployed.21

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s ManualCHAPTER 5Exercise 5.11.(a) (5.2)(b) (5.2)(f) (5.1)(g) (5.2)2.(a) p q3.(a) Yes(c) (5.3)(b) p q(b) Yes(c) Yes(d) (5.3)(e) (5.3)(c) p q(d) No; v20 2v104. We get the same results as in the preceding problem.(a) Interchange row 2 and row 3 in A to get a matrix A1 . In A1 keep row 1 as is, but addrow 1 to row 2, to get A2 . In A2 , divide row 2 by 5. Then multiply the new row 2 by 3,and add the result to row 3. The resulting echelon matrix 1 51 1 A3 0 1 5 0 0 8 25contains three nonzero-rows; hence r(A) 3.(b) Interchange row 1 and row 3 in B to get a matrix B1 . In B1 , divide row 1 by 6. Thenmultiply the new row 1 by 3, and add the result to row 2, to get B2 . In B2 , multiplyrow 2 by 2, then add the new row 2 to row 3. The resulting echelon matrix 1 16 0 B3 0 1 4 0 0 0with two nonzero-rows in B3 ; hence r(B) 2. There is linear dependence in B: row 1 isequal to row 3 2(row 2). Matrix is singular.(c) Interchange row 2 and row 3 in C, to get matrix C1 . In C1 divide row 1 to 7. Thenmultiply the new row 1 by 8, and add the result to row 2, to get C2 . In C2 , multiplyrow 2 by 7/48. Then multiply the new row 2 by 1 and add the result to row 3, to get22

Chiang/Wainwright: Fundamental Methods of Mathematical EconomicsInstructor’s ManualC3 . In C3 , multiply row 3 by 2/3, to get the echelon matrix 31 67 377 C4 0 1 12 23 0 0 1 109There are three nonzero-rows in C4 ; hence r(C) 3. The question of nonsingularity isnot relevant here because C is not square.(d) interchange row 1 and row 2 in D, to get matrix D1 (This step is optional, because we canjust as well start by dividing the original row 1 by 2 to produce the desired unit elementat the left end of the row. But the interchange of rows 1 and 2 gives us simpler numbersto work with). In D1 , multiply row 1 by 2, and add the result to row 2, to get D2 .Since the last two rows of D2 , are identical, linear dependence is obvious. To produce anechelon matrix, divide row 2 in D2 by 5, and then add ( 5) times the new row 2 to row3. The resulting echelon matrix 1 1 D3 0 1 0 001 35 0095contains two nonzero-rows; hence r(D) 2. Again, the question nonsingularity is notrelevant here.5. The link is provided by the third elementary row operation. If, for instance, row 1 of a givenmatrix is equal to row 2 minus k times row 3 (showing a specific pattern of linear combination),then by adding ( 1) times row 2 and k times row 3 to row 1, we can produce a zero-row. Thisprocess involves the third elementary row operation. the usefulness of the echelon matrixtransformation lies in its systematic approach to force out zero-rows if they exist.Exercise 5.21.(a) 6(b) 0(e) 3abc a3 b3 c32. , , , ,(c) 0(f) 8xy 2x 30 .23(d) 157

Chiang/Wainwright: Fundamental Methods of Mathematical Economics3. a f Ma h i d f Mb g i Ca Ma 4. a b Mf g h Cb Mb (b) 81 23 4 5. The cofactor of element 9 is

Chiang/Wainwright: Fundamental Methods of Mathematical Economics Instructor’s Manual 5. (a) By Theorem I, any integer root must be a divisor of 6; thus there are six candidates: 1,File Size: 1MB

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