Mathematics for Economists Solutions PDF

A-PDF MERGER DEMO Answers Pamphlet for MATHEMATICS FOR ECONOMISTS Carl P. Simon Lawrence Blume W.W. Norton and Company, Inc. Table of Contents Chapter 2 One-Vari...

A-PDF MERGER DEMO Answers Pamphlet for MATHEMATICS FOR ECONOMISTS Carl P. Simon Lawrence Blume W.W. Norton and Company, Inc. Table of Contents Chapter 2 One-Variable Calculus: Foundations 1 Chapter 3 One-Variable Calculus: Applications 5 Chapter 4 One-Variable Calculus: Chain Rule 9 Chapter 5 Exponents and Logarithms 11 Chapter 6 Introduction to Linear Algebra 13 Chapter 7 Systems of Linear Equations 15 Chapter 8 Matrix Algebra 25 Chapter 9 Determinants: An Overview 41 Chapter 10 Euclidean Spaces 45 Chapter 11 Linear Independence 52 Chapter 12 Limits and Open Sets 55 Chapter 13 Functions of Several Variables 60 Chapter 14 Calculus of Several Variables 63 Chapter 15 Implicit Functions and Their Derivatives 68 Chapter 16 Quadratic Forms and Definite Matrices 77 Chapter 17 Unconstrained Optimization 82 Chapter 18 Constrained Optimization I: First Order Conditions 88 Chapter 19 Constrained Optimization II 98 Chapter 20 Homogeneous and Homothetic Functions 106 Chapter 21 Concave and Quasiconcave Functions 110 Chapter 22 Economic Applications 116 Chapter 23 Eigenvalues and Eigenvectors 125 Chapter 24 Ordinary Differential Equations: Scalar Equations 146 Chapter 25 Ordinary Differential Equations: Systems of Equations 156 Chapter 26 Determinants: The Details 166 Chapter 27 Subspaces Attached to a Matrix 174 Chapter 28 Applications of Linear Independence 181 Chapter 29 Limits and Compact Sets 182 Chapter 30 Calculus of Several Variables II 188 Appendix 1 Sets, Numbers, and Proofs 193 Appendix 2 Trigonometric Functions 195 Appendix 3 Complex Numbers 199 Appendix 4 Integral Calculus 202 Appendix 5 Introduction to Probability 203 Figures 205  ANSWERS PAMPHLET 1 Chapter 2 2.1 i) y 5 3x22 is increasing everywhere, and has no local maxima or minima. See figure.* ii) y 5 22x is decreasing everywhere, and has no local maxima or minima. See figure. iii) y 5 x 2 1 1 has a global minimum of 1 at x 5 0. It is decreasing on (2`, 0) and increasing on (0, `). See figure. iv) y 5 x 3 1x is increasing everywhere, and has no local maxima or minima. See figure. p p v) y 5 x 3 2 x has a local maximum of 26 3 3 at 216 3, and a local p p minimum of 226 3 3 pat 16 3, but p no global maxima or minima. It increases on (2`, 216 3) and (16 3, `) and decreases in between. See figure. vi) y 5 |x| decreases on (2`, 0) and increases on (0, `). It has a global minimum of 0 at x 5 0. See figure. 2.2 Increasing functions include production and supply functions. Decreasing functions include demand and marginal utility. Functions with global critical points include average cost functions when a fixed cost is present, and profit functions. 2.3 1, 5, 22, 0. 2.4 a) x Þ 1; b) x. 1; c) all x; d) x Þ 61; e) 21 # x # 11; f ) 21 # x # 11, x Þ 0. 2.5 a) x Þ 1, b) all x, c) x Þ 21, 22, d) all x. 2.6 The most common functions students come up with all have the nonnegative real numbers for their domain. 2.8 a) 1, b) 21, c) 0, d) 3. 2.8 a) The general form of a linear function is f (x) 5 mx 1 b, where b is the y-intercept and m is the slope. Here m 5 2 and b 5 3, so the formula is f (x) 5 2x 1 3. b) Here m 5 23 and b 5 0, so the formula is f (x) 5 23x. *All figures are included at the back of the pamphlet. 2 MATHEMATICS FOR ECONOMISTS c) We know m but need to compute b. Here m 5 4, so the function is of the form f (x) 5 4x 1 b. When x 5 1, f (x) 5 1, so b has to solve the equation 1 5 4 ? 1 1 b. Thus, b 5 23 and f (x) 5 4x 2 3. d) Here m 5 22, so the function is of the form f (x) 5 22x 1 b. When x 5 2, f (x) 5 22, thus b has to solve the equation 22 5 22 ? 2 1 b, so b 5 2 and f (x) 5 22x 1 2. e) We need to compute m and b. Recall that given the value of f (x) at two points, m equals the change in f (x) divided by the change in x. Here m 5 (5 2 3)6 (4 2 2) 5 1. Now b solves the equation 3 5 1 ? 2 1 b, so b 5 1 and f (x) 5 x 1 1. f ) m 5 [3 2 (24)]6 (0 2 2) 5 276 2, and we are given that b 5 3, so f (x) 5 2(76 2)x 1 3. 2.9 a) The slope is the marginal revenue, that is, the rate at which revenue increases with output. b) The slope is the marginal cost, that is, the rate at which the cost of purchasing x units increases with x. c) The slope is the rate at which demand increases with price. d) The slope is the marginal propensity to consume, that is, the rate at which aggregate consumption increases with national income. e) The slope is the marginal propensity to save, that is, the rate at which aggregate savings increases with national income. 2.10 a) The slope of a secant line through points with x-values x and x 1 h is [m(x 1 h) 2 mx]6 h 5 mh6 h 5 m. b) For f (x) 5 x3 , f (x 1 h) 2 f (x) x 3 1 3x 2 h 1 3xh2 1 h3 2 x 3 lim 5 lim h→0 h h→0 h 5 lim 3x 2 1 3xh 1 h2 5 3x 2. h→0 For f (x) 5 x 4 , 4x 3 h 1 6x 2 h2 1 4xh3 1 h4 f 0(x) 5 lim h→0 h 5 lim 4x 3 1 6x 2 h 1 4xh2 1 h3 5 4x 3. h→0  ANSWERS PAMPHLET 3 p 2.11 a) 221x 2 , b) 224x 23 , c) 2(96 2)x 256 2 , d) 16 4 x, e) 6x 2 9 1 (146 5)x 236 5 2 (36 2)x 216 2 , f ) 20x 4 2 (36 2)x 216 2 , g) 4x 3 1 9x 2 1 6x 1 3, p h) (16 2)(x 216 2 2 x 236 2 )(4x 5 2 3 x) 1 (x 16 2 1 x 216 2 )(20x 4 2 (36 2)x 216 2 ), i) 26 (x 1 1)2 , j) (1 2 x 2 )6 (1 1 x 2 )2 , k) 7(x 5 2 3x 2 )6 (5x 4 2 6x), l) (106 3)(x 5 2 6x 2 1 3x)216 3 (5x 4 2 12x 1 3), m) 3(3x 2 1 2)(x 3 1 2x)2 (4x 1 5)2 1 8(x 3 1 2x)3 (4x 1 5). 2.12 a) The slope of the tangent line l(x) 5 mx 1 b to the graph of f (x) at x0 is m 5 f 0(x0 ) 5 2x0 5 6. The tangent line goes through the point (x0 , f (x0 )) 5 (3, 9), so b solves 9 5 6 ? 3 1 b. Thus b 5 29 and l(x) 5 6x 2 9. b) Applying the quotient rule, f 0(x) 5 (2 2 x 2 )6 (x 2 1 2)2. Evaluating this at x0 5 1, m 5 16 9. The tangent line goes through the point (1, 16 3). Solving for b, l(x) 5 (16 9)x 1 26 9. ¡ ¢ 0 ( f (x0 1 h) 1 g(x0 1 h) 2 f (x0 ) 1 g(x0 ) 2.13 ( f 1 g) (x0 ) 5 lim h→0 h f (x0 1 h) 2 f (x0 ) g(x0 1 h) 2 g(x0 ) 5 lim 1 lim h→0 h h→0 h 5 f 0(x0 ) 1 g 0(x0 ) and similarly for ( f 2 g) 0(x0 ). k f (x0 1 h) 2 k f (x0 ) (k f ) 0(x0 ) 5 lim h→0 h f (x0 1 h) 2 f (x0 ) 5 k lim 5 k f 0(x0 ). h→0 h 2.14 Let F(x) 5 x 2k 5 16 x k. Apply the quotient rule with f (x) 5 1 and g(x) 5 x k. Then f 0(x0 ) 5 0, g 0(x0 ) 5 kx0k21 , and F 0(x0 ) 5 2kx0k21 6 x02k 5 2kx02k21. 2.15 For positive x, |x| 5 x, so its derivative is 1. For negative x, |x| 5 2x, so its derivative is 21. 4 MATHEMATICS FOR ECONOMISTS ½ 2x if x. 0, 2.16, 17 a) f 0(x) 5 22x if x , 0. As x converges to 0 both from above and below, f 0(x) converges to 0, so the function is C 1. See figure. b) This function is not continuous (and thus not differentiable). As x con- verges to 0 from above, f (x) tends to 1, whereas x tends to 0 from below, f (x) converges to 21. See figure. c) This function is continuous, since limx→1 f (x) 5 1 no matter how the ± But it is not differentiable at x ±5 1, since limh ↓ 0 [ f (1 1 limit is taken. h) 2 f (1)] h 5 3 and limh ↑ 0 [ f (1 1 h) 2 f (1)] h 5 1. See figure. d) This function is C 1 at x 5 1. No matter which formula is used, the value for the derivative of f (x) at x 5 1 is 3. See figure. ½ f (x) 5 3x 0 2 if x , 1, 3 if x $ 1. 2.18 The interesting behavior of this function occurs in a neighborhood of x 5 0. ± Computing, [ f (0 1 h) 2 f (0)] h 5 h216 3 , which converges to 1` or 2` as h converges to 0 from above or below, respectively. Thus f (x) 5 x 26 3 is not differentiable at x 5 0. It is continuous at x 5 0, since limx→0 f (x) 5 0. This can easily be seen by plotting the function. See figure. 2.19 See figure. 2.20 a) 242x, b) 72x 24 , c) (456 4)x 276 2 , d) x 236 2 6 8, e) 6 2 (526 25)x 286 5 1 (36 4)x 236 2 , f ) 80x 3 1 (36 4)x 236 2 , g) 12x 2 1 18x 1 6, p h) (2x 236 2 6 413x 256 2 6 4)(4x 5 23 x)1(x 216 2 2x 236 2 )(20x 4 23x 216 2 6 2) 1 (x 16 2 1 x 216 2 )(80x 3 1 3x 236 2 6 4), i) 246 (x 1 1)3 , j) (2x 3 2 6x)6 (x 2 1 1)3 , k) 42(5x 4 2 6x)2 (x 5 2 3x 2 )5 1 7(20x 3 2 6)(x 5 2 3x 2 )6 , l) (2106 9)(5x 4 2 12x 1 3)2 (x 5 2 6x 2 1 3x)246 3 1 (106 3)(20x 3 2 12)(x 5 2 6x 2 1 3x)216 3 , m) 12(x 1 1)2 (4x 1 5)2 (x 2 1 2x) 1 96(x 1 1)(4x 1 5)(x 2 1 2x)2 1 6(4x 1 5)2 (x 2 1 2x)2 1 32(x 2 1 2x)3. 2.21 a) f 0(x) 5 (56 3)x 26 3 , so f (x) is C 1. But x 26 3 is not differentiable at x 5 0, so f is not C 2 at x 5 0. Everywhere else it is C `.  ANSWERS PAMPHLET 5 b) This function is a step function: f (x) 5 k for k # x , k 1 1, for every integer k. It is C ` except at integers, since it is constant on every interval (k, k 1 1). At integers it fails to be continuous. 2.22 C 0(20) 5 86, so C(21) 2 C(20) < C 0(20) ? 1 5 86. Direct calculation shows that C(21) 5 1020, so C(21) 2 C(20) 5 88. 2.23 C 0(x) 5 0.3x 2 2 0.5x 1 300. Then C(6.1) 2 C(6) < C 0(6) ? 0.1 5 30.78. 2.24 F 0(t) 5 86 (12)2. Thus F 0(0) 5 2 and the population increase over the next half-year is F 0(0) ? 0.5 5 1. p p 2.25 a) f (x) 5 x, and f 0(x) 5 16 2 x. f (50) 5 f (49) 1 f 0(49) ? (1.0) 5 7 1 16 14. b) f (x) 5 x 16 4 , and f 0(x) 5 16 (4x 36 4 ). Then f (9, 997) < f (10,000) 1 f 0(10,000) ? (23.0) 5 10 2 36 4,000 5 9.99925. c) f (x) 5 x 5 , and f 0(x) 5 5x 4. f (10.003) 5 f (10) 1 f 0(10) ? 0.003 5 100,000 1 50,000 ? 0.003 5 100,150. Chapter 3 3.1 a) f 0(x) 5 3x 2 1 3, so f 0(x) is always positive and f (x) is increasing throughout its domain. f (0) 5 0, so the graph of f passes through the origin. See figure. b) Early versions of the text have f (x) 5 x 4 2 8x 3 1 18x 2 11 here, with f 0(x) 5 4x 3 2 24x 2 1 18. This itself is a complicated function. f 00(x) 5 12x 2 2 48x. Thus, f 0(x) has critical points at x 5 0 and x 5 4. The point x 5 0 is a local maximum of f 0, and x 5 4 is a local minimum. Evaluating, f 0(x) is positive at the local max and negative at the local min. This means it crosses the x-axis three times, so the original function f has three critical points. Since f 0(x) is negative for small x and positive for large x, the critical points of f are, from smallest to largest, a local minimum, a local maximum, and a local minimum. Later versions have f (x) 5 x 4 2 8x 3 1 18x 2 2 11. Its y-intercept is at (0, 211); f 0(x) 5 4x 3 2 24x 2 1 36x 5 4x(x 2 2 6x 1 9) 5 4x(x 2 3)2. Critical points are at x 5 0, 3, i.e., (0,11) and (3,16). f 0. 0 (and f increasing) for 0 , x , ` (x Þ 3); f 0 , 0 (and f decreasing) for 2` , x , 0. See figure. c) f 0(x) 5 x 2 19. This function is always positive, so f is forever increasing. A little checking shows its root to be between 216 3 and 216 2. See figure. 6 MATHEMATICS FOR ECONOMISTS d) f 0(x) 5 7x 6 2 7, which has roots at x 5 61. The local maximum is at (21, 6), and the local minimum is at (1, 26). The function is decreasing between these two points and increasing elsewhere. The y-intercept is at (0, 0). See figure. e) f (0) 5 0; f 0(x) 5 (26 3)x 216 3. f 0(x) , 0 (and f decreasing) for x , 0; f 0(x). 0 (and f increasing) for x. 0. As x → 0, the graph of f becomes infinitely steep. See figure. f ) f 0(x) 5 12x 5 2 12x 3 5 12x 3 (x 2 2 1). The first derivative has roots at 21, 0 and 1. f 0(x) is negative for x , 21 and 0 , x , 1, and positive for 21 , x , 0 and x. 1. Thus f (x) is shaped like a w. Its three critical points are, alternately, a min, a max, and a min. Its values at the two minima are both 21, and its value at the maximum is 12. See figure. 3.2 Since f is differentiable at x0 , for small h, [ f (x0 1 h) 2 f (x0 )]6 h , 0. This means that for small positive h, f (x0 1 h) , f (x0 ) and, for small negative h, f (x0 1 h). f (x0 ). Thus, f is decreasing near x0. 3.3 a) f 00(x) 5 6x. The function is concave (concave down) on the negative reals and convex (concave up) on the positive reals. b) f 00(x) 5 12x 2 2 48x, which is negative for 0 , x , 4 and positive outside this interval. Thus f is concave on the interval (0, 4) and convex elsewhere. c) f 00(x) 5 2x, so f is concave on the negative reals and convex on the positive reals. d) f 00(x) 5 42x 5 , so f is concave on the negative reals and convex on the positive reals. e) f 00(x) 5 22x 246 3 6 9. This number is always less than 0 for x Þ 0. f is concave on (0, `) and on (2`, 0). It is not globally concave. p p f ) f 00(x) 5 60x 4 2 36x 2 , which is negative on the interval (2 36 5, 36 5), and positive outside it. Thus, f is concave on this interval and convex elsewhere. 3.4–5 See figures. 3.6 There is a single vertical asymptote at x 5 2. f 0(x) 5 216(x 1 4)6 (x 2 2)3 and f 00(x) 5 32(x 1 7)6 (x 2 2)4. Consequently there is a critical point at x 5 24, where the function takes the value 246 3. f 00(24). 0, so this is a local minimum. There is an inflection point at x 5 27. f is decreasing to the right of its asymptote and to the left of x 5 24, and increasing on (24, 2). See figure.  ANSWERS PAMPHLET 7 3.7 a) The leading monomial is x 21 , so f (x) converges to 0 as x becomes very positive or very negative. It also has vertical asymptotes at x 5 21 and x 5 1. f 0(x) 5 2(x 2 1 1)6 (x 2 2 1)2 ; so for x , 21 and x. 1, f (x) is decreasing. (In fact it behaves as 16 x.) It is also decreasing between the asymptotes. Thus, it goes from 0 to 2` as x goes from 2` to 21, from 1` to 2` as x goes from 21 to 1, and from 1` to 0 as x goes from 1 to 1`. See figure. b) f (x) behaves as 16 x for x very large and very small. That is, as |x| grows large, f (x) tends to 0. f 0(x) 5 (1 2 x 2 )6 (x 2 1 1)2 ; so there are critical points at 21 and 1. These are, respectively, a minimum withpvaluep216 2 and a maximump pwith value 16 2. Inflection points are at (2 3, 2 36 4), (0, 0), and ( 3, 36 4). See figure. c) This function has a vertical asymptote at x 5 21. The lead monomial is x 2 6 x 5 x, so in the tails it is increasing as x → 1` and decreasing as x → 2`. As x converges to 21 from below, f (x) tends to 2`; as x converges to 21 from above, f (x) converges to 1`. See figure. d) The lead monomial is x 2 6 x 2 5 1, so f (x) converges to 1 as |x| becomes large. It has vertical asymptotes at x 5 1 and x 5 21. In fact, f can be rewritten as f (x) 5 1 1 (3x 1 1)6 (x 2 2 1). Since f 0(x) 5 2(3x 2 1 2x 1 3)6 (x 2 2 1)2 , 0, f is always decreasing. So, its general shape is that of the function in part 7a. See figure. e) The lead monomial is x 2 6 x 5 x, so this function is increasing in x when |x| is large. When x is small near its vertical asymtote at x 5 0, it behaves as 16 x. f 00(x) 5 1 2 16 x 2 , which is 0 at 61. x 5 21 is a local maximum and x 5 1 is a local minimum. See figure. f ) This function is bell shaped. It is always positive, tends to 0 when |x| is large, and has a maximum at x 5 0 where it takes the value 1. See figure. 3.8 See figures. 3.9 a) No global max or min on D1 ; max at 1 and min at 2 on D2. b) No max or min on D1 ; min at 0 and max at 1 on D2. c) Min at 24, max at 22 on D1 ; min at 11, no max on D2. d) Min at 0, max at 10 on D1 ; min at 0, no max on D2. p p e) Min at 22, max at 12 on D1 ; min at 2 2, max at 1 2 on D2. f ) Min at 1, no max on D1 ; max at 21, no min on D2. g) No min or max on D1 ; max at 1 and min at 2 on D2. h) No max or min on D1 ; max at 1 and min at 5 on D2. 8 MATHEMATICS FOR ECONOMISTS 3.10 In this exercise x is the market price, which is a choice variable for the firm. p (x) 5 x(15 2 x) 2 5(15 2 x). This function is concave, and its first derivative is p 0(x) 5 22x 1 20. p 0(x) 5 0 at x 5 10. 3.11 From the information given, the demand function must be computed. The function is linear, and the slope is 21. It goes through the point (10, 10), so the function must be f (x) 5 20 2 x. Then the profit function (as a function of price) must be p (x) 5 (x 2 5)(20 2 x). p 0(x) 5 22x 1 25, so profit is maximized at x 5 12.5. 3.12 One can translate the proof of Theorem 3.4a in the text. Here is another idea. If ` is a secant line connecting (x0 , y0¡) and (x¢ 1 , y1 ) on the graph of a convex function f (x), then the set of points x, f (x) for x Ó (x0 , x1 ) lies above `. Taking limits, the graph of a convex function always lies above each tangent line (except where they touch). If f 0(x0 ) 5 0, then the tangent line is of the form ` (x) 5 f (x0 ) 5 b. Since f (x) is convex near x0 , f (x) must be at least as big as b for x near x0 , and so x0 is a min. 3.13 Suppose y0 , x0 ; f is decreasing just to the right of y0 and increasing just to the left of x0. It must change from decreasing to increasing somewhere between y0 and x0 , say at w0. Then, w0 is an interior critical point of f — contradicting the hypothesis that x0 is the only critical point of f0. 3.15 AC(x) 5 x 2 1 1 1 16 x. MC(x) 5 3x 2 1 1. MC(x0 ) 5 AC(x0 ) when 2x02 5 16 x0 , that is, at x0 5 2216 3. AC 0(x) 5 2x 2 16 x 2 , so AC(x) has a critical point at x 5 2216 3. Thus c is satisfied. MC(x) is increasing, AC(x) is convex, and the two curves intersect only once at x0 , so it must be that to the left of x0 , AC(x). MC(x), and hence to the right, AC(x) , MC(x). See figure. p p 3.16 Suppose C(x) 5 x. Then MC(x) 5 16 2 x, which is decreasing. p (x) 5 p p px 2 x. p 0(x) 5 p 2 16 2 x. The equation p 0(x) 5 0 will have a solution, but p 00(x) 5 216 4x 36 2 , which is always negative on the positive reals. Setting price equal to marginal cost gives a local min. Profit can always be increased by increasing output beyond this point. 3.17 a) Locate x p correctly at the intersection of the MR and MC curves. Revenue at the optimum is described by the area of the rectangle with height AR(x p ) and length x p. b) The rectangle with height AC(x p ) and length x p. c) The rectangle with height AR(x p ) 2 AC(xp) and length x p.  ANSWERS PAMPHLET 9 3.18 For demand curve x 5 a 2 bp, the elasticity at (a 2 bp, p) is « 5 2bp6 (a 2 bp). Then, « 5 21 ⇐⇒ bp 5 a 2 bp ⇐⇒ 2bp 5 a ⇐⇒ p 5 a6 b. F 0(p) ? p 2rkp2r 21 ? p 3.19 « 5 5 5 2r , constant. F(p) kp2r 3.20 x p and pp both increase. 3.21 The rectangle with height pp 2 AC(x p ) and length x p. 3.22 First, compute the inverse demand: p 5 a6 b 2 (16 b)x. Then revenue is R(x) 5 (a6 b)x 2 (16 b)x2 and MR(x) 5 a6 b 2 (26 b)x. MC(x) 5 2kx, so x p solves a6 b 2 (26 b)x 5 2kx. The solution is x p 5 a6 (2kb 1 2), and the price will be pp 5 2kab6 (2kb2 1 2b). Chapter 4 4.1 a) (g ◦ h)(z) 5 (5z 2 1)2 1 4, (h ◦ g)(x) 5 5x 2 1 19. b) (g ◦ h)(z) 5 (z 2 1)3 (z 1 1)3 , (h ◦ g)(x) 5 (x 3 2 1)(x 3 1 1). c, d) (g ◦ h)(z) 5 z, (h ◦ g)(x) 5 x. e) g ◦ h(z) 5 16 (z2 1 1), h ◦ g(x) 5 16 x 2 1 1. 4.2 a) Inside y 5 3x 2 1 1, outside z 5 y16 2. b) Inside y 5 16 x, outside z 5 y2 1 5y 1 4. c) Inside y 5 2x 2 7, outside z 5 cos y. d) Inside y 5 4t 1 1, outside z 5 3y. 4.3 a) (g ◦ h) 0(z) 5 2(5z 2 1)5 5 50z 2 10, (h ◦ g) 0(x) 5 5 ? 2x 5 10x. b) (g ◦ h) 0(z) 5 3[(z 2 1)(z 1 1)]2 (2z) 5 6z(z 2 1)2 (z 1 1)2 , (h ◦ g) 0(x) 5 2x 3 ? 3x 2 5 6x 5. c) (g ◦ h) 0(z) 5 1, (h ◦ g) 0(x) 5 1. d) (g ◦ h) 0(z) 5 1, (h ◦ g) 0(x) 5 1. e) (g ◦ h) 0(z) 5 22z6 (z2 1 1)2 , (h ◦ g) 0(x) 5 226 x 3. p 4.4 a) (g ◦ h) 0(x) 5 12 (3x 2 1 1)216 2 ? 6x 5 3x6 3x 2 1 1 ¡ ¢ b) (g ◦ h) 0(x) 5 [2(16 x) 1 5] 216 x 2. 10 MATHEMATICS FOR ECONOMISTS c) (g ◦ h) 0(x) 5 22 sin(2x 2 7). d) (g ◦ h) 0(t) 5 (4 log 3)34t11. 4.5 a) (g ◦ h) 0(x) 5 cos(x 4 ) ? 4x 3. b) (g ◦ h) 0(x) 5 cos(16 x) ? (216 x 2 ). p c) (g ◦ h) 0(x) 5 cos x6 (2 sin x). p p d) (g ◦ h) 0(x) 5 (cos x)6 2 x. e) (g ◦ h) 0(x) 5 (2x 1 3) exp(x 2 1 3x). f ) (g ◦ h) 0(x) 5 2x 22 exp(16 x). g) (g ◦ h) 0(x) 5 2x6 (x 2 1 4). h) (g ◦ h) 0(x) 5 4x(x 2 1 4) cos((x 2 1 4)2 ). ¡ ¢ 4.6 x 0(t) 5 2 and C 0(x) 5 12, so (d6 dt)C x(t) 5 2 ? 12 5 24. 4.8 a) g(y) 5 (y 2 6)6 3, 2` , y , 1`. b) g(y) 5 16 y 2 1, 2` , y , 1`, y Þ 0. c) The range of f (x) 5 x26 3 is the nonnegative reals, so this is the domain of the inverse. But notice that f (x) is not one-to one from R to its range. It is one-to-one if the domain of f is restricted to R1. In this case the inverse is g(y) 5 y36 2 , 0 # y , `. If the domain of f is restricted to R2 , the inverse is 2g(y). d) The graph of f (x) is a parabola with a global minimum at x 5 16 2, and is one-to-one on each side of it. Thus there will be two inverses. For a given y they pare the solutions to x 2 1 xp1 2 5 y. The two inverses are z 5 12 (211 4y 2 7) and z 5 12 (212 4y 2 7), with domain y $ 76 4. If y , 76 4, the equation has no solution; there is no value of x such that f (x) 5 y. 4.9 a) ( f 21 ) 0( f (1)) 5 16 3, 16 f 0(1) 5 16 3. b) ( f 21 ) 0(16 2) 5 24, 16 f 0(1) 5 2(1 1 1)2 5 24. c) ( f 21 ) 0 f (1) 5 36 2, 16 f 0(1) 5 36 2. d) ( f 21 ) 0 f (1) 5 16 3, 16 f 0(1) 5 16 3.  ANSWERS PAMPHLET 11 2n 4.10 To prove Theorem 4.4, let f (x) 5 x 16 5 16 x 16 n for n a positive integer. Applying the quotient rule, 0 ? x 16 n 2 1 ? (16 n)x (16 n)21 f 0(x) 5 x 26 n 5 2(16 n)x 2(16 n)21. For Theorem 4.5, the proof in the text applies to the case where both m and n are negative. To prove the remaining case, let f (x) 5 x 2m6 n where m, n are positive integers. Applying the quotient rule, 0 ? x m6 n 2 1 ? (m6 n)x (m6 n)21 f 0(x) 5 x 2m6 n 5 2(m6 n)x (m6 n)212(2m6 n) 5 2(m6 n)x 2(m6 n)21. Chapter 5 5.1 a) 8, b) 16 8, c) 2, d) 4, e) 16 4, f ) 1, g) 16 32, h) 125, i) 16 3125. 5.2 See figures. 5.3 By calculator to three decimal places: a) 2.699, b) 0.699, c) 3.091, d) 0.434, e) 3.401, f ) 4.605, g) 1.099, h) 1.145. See figures. 5.4 a) 1, b) 23, c) 9, d) 3, e) 2, f ) 21, g) 2, h) 16 2, i) 0. 5.5 a) 2e6x 5 18 =⇒ e6x 5 9 =⇒ 6x 5 ln 9 =⇒ x 5 (ln 9)6 6; 2 b) ex 5 1 =⇒ x 2 5 ln 1 5 0 =⇒ x 5 0; c) 2x 5 e5 =⇒ ln 2x 5 ln e5 =⇒ x ln 2 5 5 =⇒ x 5 56 (ln 2); d) 2 1 ln 56 ln 2; e) e56 2 ; f ) 5. 5.6 Solve 3A 5 A exp rt. Dividing out A and taking logs, ln 3 5 rt, and t 5 ln 36 r. 5.7 Solve 600 5 500 exp(0.05t). t 5 ln(66 5)6 0.05 5 3.65. 12 MATHEMATICS FOR ECONOMISTS 5.8 a) f 0(x) 5 e3x 1 3xe3x , f 00(x) 5 6e3x 1 9xe3x. f 0(x) 5 (2x 1 3)ex 13x22 , f 00(x) 5 (4x 2 1 12x 1 11)ex 13x22. 2 2 b) c) f 0(x) 5 8x 3 6 (x 4 1 2), f 00(x) 5 (48x 2 2 8x 6 )6 (x 4 1 2)2. d) f 0(x) 5 (1 2 x)6 ex , f 00(x) 5 (x 2 2)6 ex. e) f 0(x) 5 16 (ln x) 2 16 [(ln x)]2 , f 00(x) 5 26 [x(ln x)3 ] 2 16 [x(ln x)2 ]. f) f 0(x) 5 (1 2 ln x)6 x 2 , f 00(x) 5 (2 ln x 2 3)6 x 3. g) f 0(x) 5 x6 (x 2 1 4), f 00(x) 5 (2x 2 1 4)6 (x 2 1 4)2. 5.9 a) f (x) 5 xex =⇒ f 0(x) 5 ex (x 1 1) is positive ( f increasing) for x. 21 and negative ( f decreasing) for x , 21. f 00(x) 5 ex (x 1 2) is positive ( f convex) for x. 22, and negative ( f concave) for x , 22. As x tends to 2`, f (x) goes to 0, and as x gets large, f (x) behaves as ex. Thus, the function has a horizontal asymptote at 0 as x → 2`, grows unboundedly as x → 1`, has a global minimum at 21, and has an inflection point at 22. See figure. b) y 5 xe2x ; y 0 5 (1 2 x)e2x is positive ( f increasing) for x , 1, and negative ( f decreasing) for x. 1. y 00 5 (x 2 2)e2x is positive ( f convex) for x. 2, and negative ( f concave) for x , 2. As x gets large, f (x) tends to 0. As x goes to 2`, so does f (x). The inflection point is at x 5 2. See figure. c) y 5 12 (ex 1 e2x ); y 0 5 12 (ex 2 e2x ) is positive (y increasing) if x. 0, and negative (y decreasing) if x , 0. y 00 5 y is always positive (y always convex). See figure. 5.10 Let f (x) 5 Log(x), and let g(x) 5 10 f (x) 5 x. Then, by the Chain Rule and Theorem 5.3, g 0(x) 5 (ln 10)10 f (x) f 0(x) 5 1. 1 1 So, f 0(x) 5 5. (ln 10)10Log(x) x ln 10 5.11 The present value of the first option is 2156 (1.1)2 5 177.69. The present value of the second option is 1006 (1.1) 1 1006 (1.1)2 5 173.56. The present value of the third option is 100 1 956 (1.1)2 5 178.51. 5.12 By equation (14), the present value of the 5-year annuity is 1 2 e20.5 500 5 1870.62. e0.1 2 1 Equation (15) gives the present value of the infinitely lived annuity: 5006 (e0.1 2 1) 5 4754.17.  ANSWERS PAMPHLET 13 p 5.13 ln B(t) 5 t ln 2. Differentiating, B 0(t) ln 2 5 p. B(t) 2 t The solution to B 0(t)6 B(t) 5 r is t 5 100(ln 2)2 5 48.05. p p 5.14 ln V(t) 5 K 1 t. Differentiating, V 0(t)6 V(t) 5 16 2( t). The solution to V 0(t)6 V(t) 5 r is t 5 16 (4r 2 ), which is independent of K. A check of the second order conditions shows this to be a maximum. 5.15 ln V(t) 5 ln 2000 1 t 16 4. Differentiating, we find that V 0(t) 1 5 t 236 4. V(t) 4 The solution to V 0(t)6 V(t) 5 r is t 5 (4r )246 3. When r 5 0.1, t 5 3.39. 5.16 a) ln f (x) 5 1 2 ln(x 2 1 1) 2 12 ln(x 2 1 4). Differentiating, f 0(x) x x 5 2 2 2. f (x) x 11 x 14 µ ¶ x x So, f 0(x) 5 f (x) 2 2 2 x 11 x 14 3x 5 (x 2 1 1)16 2 (x 2 1 4)36 2. b) ln f (x) 5 2x 2 ln x. Differentiating, f 0(x)6 f (x) 5 4x ln x 1 2x 5 2x(1 1 ln x 2 ), so f 0(x) 5 2x(1 1 ln x 2 )(x 2 )x. 2 5.17 Let h(x) 5 f (x)g(x), so ln h(x) 5 ln f (x) 1 ln g(x). Differentiating, h 0(x)6 h(x) 5 f 0(x)6 f (x) 1 g 0(x)6 g(x). Multiplying both sides of the equality by x proves the claim. Chapter 6 6.1 S 5 0.05(100,000) F 5 0.4(100,000 2 S). 14 MATHEMATICS FOR ECONOMISTS Multiplying out the system gives S 5 5,000 F 1 0.4S 5 40,000. Thus S 5 5,000 and F 5 38,000, and after-tax profits are $57,000. Including contributions, after-tax profits were calculated to be $53,605, so the $5,956 contribution really cost only $57, 000 2 $53, 605 5 $3, 395. 6.2 Now S 5 0.05(100,000 2 C 2 F), so the equations become C 1 0.1S 1 0.1F 5 10,000 0.05C 1 S 1 0.05F 5 5,000 0.4C 1 0.4S 1 F 5 40,000. The solution is C 5 6,070, S 5 2,875, and F 5 36,422. 6.3 x1 5 0.5x1 1 0.5x2 1 1, x2 5 0x1 1 0.25x2 1 3. The solution is x1 5 6, x2 5 4. 6.4 Solving the system of equations x1 5 0.5x1 1 0.5x2 1 1 and x2 5 0.875x1 1 0.25x2 1 3 gives x1 5 236 and x2 5 238; this is infeasible. 6.5 0.002 ? 0.9 1 0.864 ? 0.1 5 0.0882, and 0.004 ? 0.8 1 0.898 ? 0.2 5 0.1828. ½ ¾ xt11 5 0.993xt 1 0.106yt 6.6 For black females,. yt11 5 0.007xt 1 0.894yt To find the stationary distribution, set xt11 5 xt 5 x and yt11 5 yt 5 y: x 5 0.9381 and y 5½ 0.0619. ¾ xt11 5 0.997xt 1 0.151yt For white females,. yt11 5 0.003xt 1 0.849yt Stationary solution: x 5 0.9805 and y 5 0.0195. 6.7 The equation system is 0.16Y 2 1500r 5 0 0.2Y 1 2000r 5 1000. The solution is r 5 0.2581 and Y 5 2419.35.  ANSWERS PAMPHLET 15 Chapter 7 7.1 a and e. 7.2 a) The solution is x 5 5, y 5 6, z 5 2. b) The solution is x1 5 1, x2 5 22, x3 5 1. 7.3 a) x 5 176 3, y 5 2136 3. b) x 5 2, y 5 1, z 5 3. c) x 5 1, y 5 21, z 5 22. 7.4 Start with system (*): a11 x1 1 ? ? ? 1 a1n xn 5 b1...... ai1 x1 1 ? ? ? 1 ain xn 5 bi...... a j1 x1 1 ? ? ? 1 a jn xn 5 bj...... an1 x1 1 ? ? ? 1 ann xn 5 bn. 1) Change system (*) to a11 x1 1 ? ? ? 1 a1n xn 5 b1...... ai1 x1 1 ? ? ? 1 ain xn 5 bi...... (rai1 1 a j1 )x1 1 ? ? ? 1 (rain 1 a jn )xn 5 (rbi 1 bj )...... an1 x1 1 ? ? ? 1 ann xn 5 bn. 2) Change the ith equation of system (*) to rai1 x1 1 ? ? ? 1 rain xn 5 rbi. 16 MATHEMATICS FOR ECONOMISTS 3) Change system (*) to a11 x1 1 ? ? ? 1 a1n xn 5 b1...... a j1 x1 1 ? ? ? 1 a jn xn 5 bj...... ai1 x1 1 ? ? ? 1 ain xn 5 bi...... an1 x1 1 ? ? ? 1 ann xn 5 bn. Reverse operations: 1) Subtract r times equation i from equation j, leaving other n 2 1 equations intact. 2) Multiply the ith equation through by 16 r. 3) Interchange the ith and jth equations again. 7.5 The system to solve is 0.20Y 1 2000r 5 1000 0.16Y 2 1500r 5 0. Solving the second equation for Y in terms of r gives Y 5 9375r. Substitut- ing into the first equation, 3875r 5 1000, so r 5 0.258 and Y 5 2419.35. 7.6 a) The system to solve is: sY 1 ar 5 I 0 mY 2 hr 5 0. Solving the second equation for Y in terms of r gives Y 5 (h6 m)r. Substituting into the first equation gives (sh 1 am)r6 m 5 I 0 , so r 5 mI 0 6 (sh 1 am) and Y 5 hI 0 6 (sh 1 am).  ANSWERS PAMPHLET 17 b, c) Differentiate the solutions with respect to s: ]r hmI0 ]Y h2 I0 52 , 0 and 52 , 0. ]s (sh 1 am)2 ]s (sh 1 am)2 7.7 Solving for y in terms of x in the second equation gives y 5 2x 2 10. Substituting this into the first equation gives a new equation that must be satisfied by all solutions. This equation is 230 5 4. Since this is never satisfied, there are no solutions to the equation system. 7.8 If a22 Þ 0, then x2 5 (b2 2 a21 x1 )6 a22. Substituting into the first equation gives a11 2 a12 a21 a12 b1 5 x1 1 b2 a22 a22 a22 b1 2 a12 b2 x1 5. a11 a22 2 a12 a21 A similar calculation solving the first equation for x1 ends up at the same point if a21 Þ 0. The division that gives this answer is possible only if a11 a22 2 a12 a21 Þ 0. In this case, a11 b2 2 a21 b1 x2 5. a11 a22 2 a12 a21 7.9 (1) Add 0.2 times row 1 to row 2. (2) Add 0.5 times row 1 to row 3. (3) Add 0.5 times row 2 to row 3.     1 0   1 2 0 µ ¶ µ ¶ 1 0 1 0 214 0 1 1 0 0.5 7.10  0 0 1  , ,  0 , , 0 1. 0 1 6 0 0 1 0.3 0 0 0 0 0 0 0 7.11 a) The original system, the row echelon form, and the reduced row echelon form are, respectively, µ ¶ µ ¶ µ ¶ 3 3 | 4 3 3 | 4 1 0 | 176 3 , ,. 1 21 | 10 0 22 | 266 3 0 1 | 2136 3 The solution is x 5 176 3, y 5 2136 3. 18 MATHEMATICS FOR ECONOMISTS b) The original system, the row echelon form, and the reduced row echelon form are, respectively,     4 2 23 | 1 4 2 23 | 1 6 3 25 | 0  ,  0 16 2 116 4 | 356 4  , 1 1 2 | 9 0 0 216 2 | 236 2   1 0 0 | 2 0 1 0 | 1. 0 0 1 | 3 The solution is x 5 2, y 5 1, z 5 3. c) The original system, the row echelon form, and the reduced row echelon form are, respectively,     2 2 21 | 2 2 2 21 | 2 1 1 1 | 22  ,  0 26 4 | 22  , 2 24 3 | 0 0 0 36 2 | 23   1 0 0 | 1 0 1 0 | 21 . 0 0 1 | 22 The solution is x 5 1, y 5 21, z 5 22. 7.12 The original system, the row echelon form, and the reduced row echelon form are, respectively,     1 1 3 22 | 0 1 1 3 22 | 0  2  3 7 22 | 9, 0  1 1 2 | 9,  3 5 13 29 | 1 0 0 2 27 | 217  22 1 0 21 | 0 0 0 0 216 2 | 236 2   1 0 0 0 | 21 0 1  0 0 | 1. 0 0 1 0 | 2 0 0 0 1 | 3 The solution is w 5 21, x 5 1, y 5 2, z 5 3. µ ¶ µ ¶ µ ¶ µ ¶ 1 1 1 0 1 3 4 1 0 1 7.13 a) and , b) and , 0 1 0 1 0 21 21 0 1 1     21 1 1 0 c)  0 21  and  0 1 . 0 0 0 0  ANSWERS PAMPHLET 19 7.14 The original system and the reduced row echelon form are, respectively, µ ¶ à ! 24 6 4 | 4 1 0 5 4 | 5 4 and. 2 21 1 | 1 0 1 3 | 3 2 2 The solution set is the set of all (x, y, z) triples such that x 5 5 4 2 5 4 z and y 5 32 2 32 z as z ranges over all the real numbers. 7.15 The original system and the row echelon form are, respectively, µ ¶ µ ¶ 1 1 | 1 1 1 | 1 and. 1 2k | 1 0 2(k 1 1) | 0 If k 5 21, the second equation is a multiple of the first. In the row echelon form this appears as the second equation 0 1 0 5 0. Any solution to the first equation solves the second equation as well, and so there are infinitely many solutions. For all other values of k there is a unique solution, with x1 5 1 and x2 5 0. 7.16 a) The original system and the reduced row echelon form are, respectively,     1 0 0 3 | 12 1 2 1 21 | 1  11 11   3 21 21 2 | 3 0 1 0 2 11 1 | 2 11 4    and  .  0 21 1 21 | 1   0 0 1 2 12 11 | 7 11  2 3 3 23 | 3 0 0 0 0 | 0 The variable z is free and the rest are basic. The solution is w5 12 11 2 3 11 z x 5 2 11 4 1 1 11 z y5 7 11 1 12 11 z. b) The original system and the reduced row echelon form are, respectively,     1 21 3 21 | 0 1 0 11 5 2 35 | 3 5 1 21 | 3    4 1  0 1 2 45 2 | 3  3 and  . 6 5 5 7 1 1 | 0 0 0 0 | 0 3 2 5 21 | 3 0 0 0 0 | 0 20 MATHEMATICS FOR ECONOMISTS The variables y and z are free, while w and x are basic. The solution is w5 3 5 2 11 5 y 1 35 z x5 3 5 1 45 y 2 25 z. c) The original system and the reduced row echelon form are, respectively,     1 2 3 21 | 1 1 0 0 0 | 176 65  21 1 2 3 | 2 0 1 0 0 | 76 65    and  .  3 21 1 2 | 2 0 0 1 0 | 226 65  2 3 21 1 | 1 0 0 0 1 | 326 65 All variables are basic. There are no free variables. The solution is w 5 176 65, x 5 76 65, y 5 226 65, z 5 326 65. d) The original system and the reduced row echelon form are, respectively,     1 1 21 2 | 3 1 1 21 2 | 3  2 2 22 4 | 6 0 0 0 0 | 0   and  .  23 23 3 26 | 29  0 0 0 0 | 0 22 22 2 24 | 26 0 0 0 0 | 0 Variable w is basic and the remaining variables are free. The solution is w 5 3 2 x 1 y 2 2z. 7.17 a) The original system and the reduced row echelon form are, respectively, µ ¶ µ ¶ 1 1 1 | 13 1 0 2 54 | 1 and. 1 5 10 | 61 0 1 9 4 | 12 To have x and y integers, z should be an even multiple of 2, i.e., 4, 8, 16,.... To have y $ 0, z # 166 3. So, z 5 4, x 5 6, y 5 3. b) 4 pennies, 6 nickels, 6 dimes! 16 coins worth 94 cents. 7.18 The reduced row echelon form of the system is   1 0 | 1 0 1 | 1 . 0 0 | 81a The last equation has solutions only when a 5 28. In this case x 5 y 5 1.  ANSWERS PAMPHLET 21 7.19 a) The second equation is 21 times the first equation. When the system is row-reduced, the second equation becomes 0x 1 0y 5 0; that is, it is redundant. The resulting system is   1 1 | 1 0 1 2 q 1 p | 1 2 q. 0 0 | 0 This system has no solution if and only if 1 2 q 1 p 5 0 and 1 2 q Þ 0. This happens if and only if p 5 2(1 2 q). With the nonnegativity constraints p, q $ 0, this can never happen unless q. 1. So the equation system always has a solution. If q 5 1 and p 5 0, the equation system has infinitely many solutions with x 5 1 2 y; otherwise it has a unique solution. b) If q 5 2 and p 5 1, the system contains the two equations x 1 y 5 1 and x 1 y 5 0, which cannot simultaneously be satisfied. More generally, if q Þ 1 and p 5 q 2 1, the equation system has no solution. 7.20 a) A row echelon form of this matrix is µ ¶ 2 24 , 0 0 so its rank is 1. b) A row echelon form of this matrix is µ ¶ 2 24 2 , 0 0 2 so its rank is 2. c) A row echelon form of this matrix is   1 6 27 3 0 3 1 1, 0 0 0 2 so its rank is 3. d) A row echelon form of this matrix is   1 6 27 3 5 0 3 1 1 4  , 0 0 0 2 1 0 0 0 0 0 so its rank is 3. 22 MATHEMATICS FOR ECONOMISTS e) A row echelon form of this matrix is   1 6 27 3 1 0 3 1 1 1, 0 0 0 2 5 so its rank is 3. 7.21 a) i) Rank M 5 #rows 5 #cols, so there is a unique solution (0, 0). ii) Rank M 5 #rows , #cols, so there are infinitely many solutions. iii) Rank M 5 #cols, so there is a unique solution (0, 0). iv) Rank M 5 #rows 5 #cols, so there is a unique solution (0, 0, 0). v) Rank M , #rows 5 #cols, so there are infinitely many solutions. b) i) Rank M 5 #rows 5 #cols, so there is a unique solution. ii) Rank M 5 #rows , #cols, so there are infinitely many solutions. iii) Rank M 5 #cols, so there are either zero solutions or one solution. iv) Rank M 5 #rows 5 #cols, so there is a unique solution. v) Rank M , #rows 5 #cols, so there are zero or infinitely many solutions. 7.22 a) Rank M 5 1 , #rows 5 #cols, so the homogeneous system has in- finitely many solutions and the general system has either 0 or infinitely many solutions. b) Rank M 5 2 5 #rows , #cols, so the homogeneous system has in- finitely many solutions and the general system has infinitely many solu- tions. c) Rank M 5 3 5 #rows , #cols, so the homogeneous system has in- finitely many solutions and the general system has infinitely many solu- tions. d) Rank M 5 3 , #rows , #cols, so the homogeneous system has in- finitely many solutions and the general system has either zero or infinitely many solutions. c) Rank M 5 3 5 #rows , #cols, so the homogeneous system has in- finitely many solutions and the general system has infinitely many solu- tions. 7.23 Checking the reduced row echelon forms, only c has no nonzero rows.  ANSWERS PAMPHLET 23 7.24 Let A be an n 3 n matrix with row echelon form R. Let a(j) be the number of leading zeros in row j of R. By definition of R, 0 # a(1) , a(2) , a(3) , ? ? ? until one reaches k so that a(k) 5 n; then a(j) 5 n for all j $ k. It follows that a(j) $ j 2 1 for all j. If A is nonsingular, a(n) , n. Since a(n) $ n 2 1, a(n) 5 n 2 1. This means a(j) 5 j 2 1 for all j, and so the jth entry in row j (diagonal entry) is not zero. Conversely, if every diagonal entry of R is not zero, a(j) , j for all j. Since a(j) $ j 2 1, a(j) 5 j 2 1 for all j. Since a(n) 5 n 2 1, A has full rank, i.e., is nonsingular. 7.25 i) The row-reduced row echelon form of the matrix for this system in the variables x, y, z, and w is µ ¶ 1 2 0 21 | 3 4. 0 0 1 0 | 1 4 The rank of the system is 2. Thus, two variables can be endogenous at any one time: z and one other. For example, the variables x and z can be solved for in terms of w and y, and the solution is x 5 36 4 2 2y 2 w and z 5 16 4. ii) The row-reduced row echelon form of the matrix for this system is   1 0 21 0 | 1 0 1 1 0 | 0. 0 0 0 1 | 0 This matrix has rank 3, so three variables can be solved for in terms of the fourth. In particular, x, y, and w can be solved for in terms of z. One solution is x 5 1 1 z, y 5 2z, and w 5 0. 7.26 C 1 0.1S 1 0.1F 2 0.1P 5 0 0.05C 1 S 2 0.05P 5 0 0.4C 1 0.4S 1 F 2 0.4P 5 0. The reduced row-echelon form for the matrix representing the system is   1 0 0 20.0595611 | 0  0 1 0 20.0470219 | 0 . 0 0 1 20.357367 | 0 24 MATHEMATICS FOR ECONOMISTS Thus the solution is C 5 0.0595611P, S 5 0.0470219P, and F 5 0.357367P. 7.27 The equation system is 0.2Y 1 2000r 1 0Ms 5 1000 0.16Y 2 1500r 2 Ms 5 2M 0. The reduced row echelon form of the matrix is µ ¶ 1 0 23.22581 | 2419.36 1 3.22581M 0. 0 1 0.000322581 | 0.258065 2 0.000322581M 0 Thus, a solution is Y 5 2419.36 1 3.22581M 0 1 3.22581Ms and r 5 0.258065 2 0.000322581M 0 2 0.000322581Ms. 7.28 a) Row reduce the matrix µ ¶ s a | I0 1 G. m 2a | Ms 2 M 0 b) The solution is h(I p 1 G) 1 a(Ms 2 M p ) Y 5 sh 1 am m(I p 1 G) 1 s(Ms 2 M p ) r5. sh 1 am c) Increases in I p , G and Ms increase Y. Increases in I p , G and M ◦ increase r. Increases in M ◦ =⇒ decreases in Y. Increases in M s =⇒ decreases in r. 7.29 a) Here is one possibility. Row reducing the matrix associated with the system gives   1 0 11 5 0 | 3 5   0 1 2 45 0 | 3 5 . 0 0 0 1 | 0 A solution is then w 5 3 5 2 11 5 y, x5 3 5 1 45 y, and z 5 0. b) If y 5 0, then a solution is w 5 35 , x 5 35 , and z 5 0.  ANSWERS PAMPHLET 25 c) Trying to solve the system in terms of z will not work. To see this, take z over to the right-hand side. The coefficient matrix for the resulting 3 3 3 system has rank 2. The system has infinitely many solutions. 7.30 The rank of the associated matrix is 2; twice the second equation plus the first equation equals the third equation. The reduced row echelon form is   1 0 11 5 2 35 | 3 5   0 1 2 45 2 5 | 3 5 . 0 0 0 0 | 0 In this case w and x can be solved for in terms of y and z. However, there is no successful decomposition involving three endogenous variables because no matrix of rank 2 can have a submatrix of rank 3. Chapter 8 µ ¶ µ ¶ 2 4 0 0 3 23 8.1 a) A 1 B 5 , A 2 D undefined, 3B 5 , 4 22 4 12 23 6     µ ¶ 0 4 2 6 5 3 DC 5 , BT 5  1 21 , AT C T 5  1 10 , 4 1 21 2 5 1 µ ¶ µ ¶ 3 3 22 22 22 C1D5 , B2A5 , AB undefined, 4 0 4 0 0 µ ¶ µ ¶ 21 22 21 CE 5 , 2D 5 , (CE)T 5 ( 21 4 ), 4 21 21 µ ¶ µ ¶ 1 21 2 1 5 B 1 C undefined, D 2 C 5 , CA 5 , 22 2 6 10 1   2 6 EC undefined, (CA)T 5  1 10 , ET C T 5 (CE)T 5 ( 21 4 ). 5 1 µ ¶µ ¶ µ ¶ 2 1 2 3 1 4 5 4 b) DA 5 5 1 1 0 21 2 2 2 3     2 0 µ ¶ 4 2 2 1 AT DT 5  3 21  ? 5  5 2  5 (DA)T. 1 1 1 2 4 3 µ ¶ µ ¶ 4 3 5 3 c) CD 5 , DC 5. 5 2 4 1 8.3 If A is 2 3 2 and B is 2 3 3, then AB is 2 3 3, so B T AT 5 (AB)T is 3 3 2. But AT is 2 3 2 and BT is 3 3 2, so AT BT is not defined. 26 MATHEMATICS FOR ECONOMISTS µ ¶ 2 25 8.5 a) AB 5 5 BA. 25 2 µ ¶µ ¶ µ ¶ r 0 a b ra 1 0c rb 1 0d b) 5 , 0 r c d 0a 1 rc 0b 1 rd µ ¶µ ¶ µ ¶ a b r 0 ar 1 b0 a0 1 br 5. c d 0 r cr 1 d0 c0 1 dr µ ¶ ra rb Both equal. More generally, if B 5 rI then AB 5 A(rI) 5 rc rd r (AI) 5 rA. BA 5 (rI)A 5 r (IA) 5 rA, too. 8.6 The 3 3 3 identity matrix is an example of everything except a row matrix and a column matrix. The book gives examples of each of these. µ ¶ µ ¶µ ¶ 21 221 2 21 2 8.7 5 , and 21 221 2 21 2 µ ¶µ ¶ µ ¶ 3 6 3 6 3 6 5. 21 22 21 22 21 22 8.8 a) Suppose that U 1 and U 2 are upper triangular; i.e., each Uijk 5 0 for i. j. Then, [U 1 1 U 2 ]ij 5 Uij1 1 Uij2 5 0 if i. j. For multiplication, the (i, j)th entry of U 1 U 2 is X X [U 1 U 2 ]ij 5 Uik1 Ukj2 1 Uik1 Ukj2. k,i k$i The first term is 0 because U 1 is upper triangular. If i. j, the second term is 0 because U 2 is upper triangular. Thus, if i. j, [U 1 U 2 ]ij 5 0, and so the product is upper triangular. If L1 and L2 are lower triangular, then (L1 )T and (L2 )T are upper triangular. By the previous paragraph, (L1 )T 1 (L2 )T is upper triangular, and so L1 1 L2 5 [(L1 )T 1 (L2 )T ]T is lower triangular. Similarly, L1 L2 5 [(L2 )T (L1 )T ]T is lower triangular. If D is both lower and upper triangular, and if i. j, Dij 5 0 (lower) and D ji 5 0 (upper), so D is diagonal. Conversely, if D is diagonal, it is obviously both upper and lower triangular. Consequently, if D1 and D2 are diagonal, then D1 1D2 and D1 D2 are both upper and lower triangular, and hence diagonal. b) Clearly, D , U; so D > U 5 D. If M is a matrix in S > U, then for i , j, Mij 5 0 (upper). Thus, M ji 5 0 (symmetric), so M is diagonal. If M is diagonal, then for i Þ j, Mij 5 0 5 M ji ; so M is symmetric. Hence D , S.  ANSWERS PAMPHLET 27    c) a1 0 ? ? ? 0 b1 0 ??? 0  0 a2 ? ? ? 0   0 b2 ??? 0     ........  .......  ....  .....  0 0 ? ? ? an 0 0 ? ? ? bn   a1 b1 0 ??? 0  0 a2 b2 ? ? ? 0   5........  .... 0 0 ??? an bn   b1 a1 0 ??? 0  0 b2 a2 ? ? ? 0   5........  .... 0 0 ? ? ? bn an    b1 0 ??? 0 a1 0 ? ? ? 0  ??? 0    0 b2   0 a2 ? ? ? 0  5........  ...... .. .........  0 0 ? ? ? bn 0 0 ? ? ? an (This also shows D is closed under multiplication.) Not true for U. For example, µ ¶µ ¶ µ ¶ 1 0 4 0 4 0 5 ; 2 3 5 6 23 18 µ ¶µ ¶ µ ¶ 4 0 1 0 4 0 5. 5 6 2 3 17 18 µ ¶ a b Symmetric matrices generally do not commute. Let A 5 and µ ¶ b c d e B5. Then, (AB)12 5 ae 1 bf and (BA)12 5 bd 1 ec. These e f two terms are generally not equal. 8.9 There are n choices for where to put the 1 in the first row, n 2 1 choices for where to put the one in the second row, etc. There are n ? (n 2 1) ? ? ? 1 5 n! permutation matrices. 8.10 Not closed under addition: The identity matrix is a permutation matrix, but I 1 I 5 2I is not. Closed under multiplication: Suppose P and Q are two n 3 n permutation matrices. First, show that each row of PQ has exactly one 1 and n 2 1 0s in it. The entries in row i of PQ are calculated by multiplying row i of P by the 28 MATHEMATICS FOR ECONOMISTS various columns of Q. If Pij 5 1, then (PQ)ik 5 0 unless column k of Q has its 1 in row j. Since Q is a permutation matrix, one and only one column of Q has a 1 in row j. So, there is one k such that (PQ)ik 5 1 and n 2 1 k’s with (PQ)ik 5 0; that is, row i of PQ has one 1 and n 2 1 0s. The transpose of a permutation matrix is a permutation matrix. So the same argument shows that each row of QT P T has one 1 and n 2 1 0s. But each row of QT P T is a column of PQ. So, every row and every column of PQ contains only one 1 and n 2 1 0s. Thus, PQ is a permutation matrix. 8.12 The three kinds of elementary n 3 n matrices are the Eij ’s, the Ei (r )’s, and the Eij (r )’s in the notation of this section. Theorem 8.2 gives the proof for the Eij ’s. For the Ei (r )’s, a generic element ehj of Ei (r ) is   ehj 5 0 if h Þ j, e 5 1 if h Þ i,  hh eii 5 r. The (k, m)th entry of Ei (r ) ? A is X n ½ akm if k Þ i ekj ? a jm 5 ekk akm 5 j51 rakm if k 5 i. So, Ei (r ) ? A is A with its ith row multiplied by r. We now work with Eij (r ), the result of adding r times row i to row j in the identity matrix I. The only nonzero entry in row i is the 1 in column i. So row j of Eij (r ) has an r in column i, in addition to the 1 in column j. In symbols, ehh 5 1 for all h e ji 5 r. ehk 5 0 for h Þ k and (h, k) Þ (j, i). Since the elements in the hth row of Eij (r ) ? A are the products of row h of Eij (r ) and the columns of A, rows of Eij (r ) ? A are the same as the rows of A, except for row j. The typical mth entry in row j of Eij (r ) ? A is X n e jk ? akm 5 e jj a jm 1 e ji aim 5 a jm 1 raim , k51 since the other e jk ’s are zero. But this states that row j of Eij (r ) ? A is (row j of A) 1 r (row i of A).  ANSWERS PAMPHLET 29 8.13 We saw in Chapter 7 that by using a finite sequence of elementary row operations, one can transform any matrix A to its (reduced) row echelon form (RREF) U. Suppose we apply row operations R1 ,..., Rm in that order to reduce A to RREF U. By Theorem 8.3, the same affect can be achieved by premultiplying A by the corresponding elementary matrices E1 ,... , Em so that Em ? Em21 ? ? ? E2 ? E1 ? A 5 U. Since U is in echelon form, each row has more leading zeros than its prede- cessor; i.e., U is upper triangular. 8.14 a) Permutation matrix P arises by permuting the rows of the m 3 m identity matrix I according to the permutation s : h1,... , mj → h1,... , mj, so that row i of P is row s(i) of I: ½ 1 if j 5 s(i) pij 5 0 otherwise. The (i, k)th entry of PA is: X m pij a jk 5 pis(i) as(i)k 5 as(i)k , j51 the (s(i), k)th entry of A. Row i of PA is row s(i) of A. b) AP 5 [[AP]T ]T 5 [P T AT ]T. If Pij 5 1, then PjiT 5 1. Applying part a shows that [AP]Tjk 5 ATik , so APkj 5 Aki. 8.15 Carry out the multiplication. In the first case, µ ¶µ ¶ µ ¶µ ¶ µ ¶ 2 1 1 21 1 21 2 1 1 0 5 5. 1 1 21 2 21 2 1 1 0 1 The computation for the second case is carried out in a similar fashion. 8.16 Carry out the multiplication. µ ¶ µ ¶ 8.17 | 1 0 0 b c d | 0 1 → | 0 1 c d 0 b | 1 0 µ ¶ µ ¶ 1 d6 c | 0 16 c 1 0 | 2d6 bc 16 c → →. 0 1 | 16 b 0 0 1 | 16 b 0 30 MATHEMATICS FOR ECONOMISTS Since a 5 0,  d 1  d b  2 2  bc c  ad 2 bc ad 2 bc   5 c a . 1 0 2 b ad 2 bc ad 2 bc 8.18 Carry out the multiplication.   ..  8.19 a).. 1 1  2 1. 1 0  → 1 2. 2 0.... 1 1. 0 1 1 1. 0 1 .   .. 1 12.. 1 0 1 0 1 1 →.. 2 →... 0 12.. 2 12 1 0 1. 21 2 µ ¶ 1 1 The inverse is. 21 2 µ ¶ 46 6 256 6 b) The inverse is. 226 6 46 6 c) Singular.   .  d). 2 4 0.. 1 0 0 1 2 0.. 1 0 0    2 .. → 0..   4 6 3. 0 1 0  22 3. 22 1 0  .... 26 210 0. 0 0 1 0 2 0. 3 0 1 .. 1  ..  1 2 0. 2 0 0 1 0 0. 2 52 0 21     .  .  →  0 1 2 3.. 1 2 1 0  →  0 1 0.. 3 1 .  2 2   2 0 2 .. 1. 0 0 1. 1 1 0 3 3 3 0 1.. 1 1 3 1 3 3   2 52 0 21   e) The inverse is  3 2 0 1 2 . 1 1 1 3 3 3   2 9 2 2 15 2 11 2  1   3 2 73 13 2 83  f ) The inverse is   1 3 .   24 3 4 21 3 4  21 1 21 1  ANSWERS PAMPHLET 31 µ ¶µ ¶ µ ¶ 1 21 5 2 8.20 a) A21 5 5. 21 2 3 1      26 36 2 21 4 3 b) A21 5  13 23 2   20  5  22 . 56 2 216 3 16 3 3 1      256 2 0 21 2 1 c) A21 5  36 2 0 16 2   1  5  0 . 16 3 16 3 16 3 26 21 8.21 A n 3 n and AB defined implies B has n rows. A n 3 n and BA defined implies B has n columns. µ ¶ µ ¶ µ ¶ 34 21 13 8 22 2 23 8.22 A 5 4 ,A 5 3 ,A 5. 21 13 8 5 23 5 8.23 To prove that Eij ? Eij 5 I, write the (h, k)th entry of Eij as   1 if h 5 i, k 5 j,  1 if h 5 j, k 5 i, ehk 5  1 if h Þ i, j and h 5 k, 0 otherwise. Let ahk denote the (h, k)th entry of Eij ? Eij :  X n  ehj e jk if h 5 i, ahk 5 ehr erk 5 ehi eik if h 5 j,  r 51 ehh ehk if h Þ i, j. If h 5 i, case 1 tells us that ½ 0 if k Þ h ahk 5 1 if k 5 h. If h 5 j, case 2 tells us that ½ 0 if h Þ k ahk 5 1 if h 5 k. If h Þ i, j, case 3 tells us that ½ 0 if h Þ k ahk 5 1 if h 5 k. In other words, (ahk ) is the identity matrix. 32 MATHEMATICS FOR ECONOMISTS To see that Ei (r ) ? Ei (16 r ) 5 I, one easily checks that the inverse of diagha1 , a2 ,... , an j is diagh16 a1 ,... , 16 an j, where the entries listed are the diagonal entries of the diagonal matrix. To see that Eij (r ) ? Eij (2r ) 5 I, write ehk for the (h, k)th entry of Eij (r ):   1 if h 5 k, ehk 5 r if (h, k) 5 (j, i),  0 otherwise, as in Exercise 8.12. Let fhk be the (h, k)th entry of Eij (2r ), with 2r replacing r in case 2. Then, the (h, k)th entry of Eij (r ) ? Eij (2r ) is X ½ n ehh ? fhk if h Þ j ahk 5 ehl flk 5 e jj f jk 1 e ji fik if h 5 j. l51 ½ 1 if h 5 k If h Þ j, ahk 5 ehh fhk 5 0 if h Þ k. ½ e jj f ji 1 e ji fii 5 2r 1 r 5 0 if k Þ j If h 5 j, ahk 5 e jj f jj 1 e ji fij 5 1 ? 1 1 r ? 0 5 1 if k 5 j. ½ 1 if h 5 k So, ahk 5 0 if h Þ k, and Eij (r ) ? Eij (2r ) 5 I. µ ¶ a 0 8.24 a) is invertible ⇐⇒ ad 2 bc 5 ad Þ 0 ⇐⇒ a Þ 0 and d Þ 0. c d µ ¶ µ ¶ a 0 21 1 d 0 b) 5 , lower triangular. c d ad 2c a µ ¶ µ ¶ a b 21 1 d 2b c) 5 , upper triangular. 0 d ad 0 a 8.25 a) Part a holds since A21 ? A 5 A ? A21 5 I implies that A is the inverse of A21. To prove b, compute that I 5 I T 5 (AA21 )T 5 (A21 )T ?AT. So, (AT )21 5 (A21 )T. To prove c, observe that (AB) ? (B21 A21 ) 5 A(BB 21 )A21 5 A ? I ? A21 5 A ? A21 5 I. Similarly, (B21 A21 ) ? (AB) 5 I.  ANSWERS PAMPHLET 33 b) Since (A1 ? ? ? Ak )(A21 21 21 k ? Ak21 ? ? ? A1 ) 5 (A1 ? ? ? Ak21 )(Ak A21 21 21 k )(Ak21 ? ? ? A1 ) 5 (A1 ? ? ? Ak21 )(A21 21 k21 ? ? ? A1 ) 5 (A1 ? ? ? Ak22 )(Ak21 ? A21 21 21 k21 )(Ak22 ? ? ? A1 ) 5 ? ? ? 5 A1 A21 1 5 I. So, (A1 ? ? ? Ak )21 5 A21 21 21 k ? Ak21 ? ? ? A1. µ ¶ µ ¶ 1 1 0 1 c) For example, A 5 1 2 and B 5 1 2 , or just take an invertible 2 0 2 1 A and let B 5 2A. 1 1 1 d) Even for 1 3 1 matrices, Þ 1 , in general. a1b a b 8.26 a) One can use the statement and/or method of Exercise 8.25b with A1 5 ? ? ? 5 Ak 5 A. b) Ar ? As 5 (A ? ? ? A) ? (A ? ? ? A) 5 A ? ? ? A{z ? A???A }. | {z } | {z } | r times s times r 1s times µ ¶ 1 21 1 c) (rA) ? A 5 r ? ? A ? A21 5 1 ? I 5 I. r r 8.27 a) Applying AB 5 BA (k 2 1) times, we easily find ABk 5 A ? BBk21 5 BAB k21 5 BAB ? Bk22 5 B2 AB k22 5 ? ? ? 5 Bk21 ? A ? B 5 Bk21 ? B ? A 5 Bk A. Use induction to prove (AB)k 5 (BA)k if AB 5 BA. It is true for k 5 1 since AB 5 BA. Assume (AB)k21 5 (BA)k21 and prove it true for k: (AB)k 5 (AB)k21 AB 5 Ak21 Bk21 AB, by inductive hypothesis 5 Ak21 AB k21 B, by first sentence in a 5A B. k k µ ¶ µ ¶ 1 1 1 0 b) Let A 5 and B 5. Then 0 1 1 1 µ ¶ ¶ µ 5 3 5 2 (AB) 5 2 but A B 5 2 2. 3 2 2 1 34 MATHEMATICS FOR ECONOMISTS More generally, suppose that A and B are non-singular. If ABAB 5 A2 B2 , then premultiplying by A21 and postmultiplying by B21 give AB 5 BA. c) (A1B)2 5 A2 1AB 1BA1B 2. (A1B)2 2(A2 12AB 1B2 ) 5 BA2AB. This equals 0 if and only if AB 5 BA.   16 d1 0 ??? 0  0 16 d2 ??? 0    8.28 D21 5 ........ . ....  0 0 ? ? ? 16 dn µ ¶21 µ ¶ a b 1 c 2b 8.29 5 , a symmetric matrix. b d ac 2 b2 2b a 8.30 Let U be an n 3 n upper-triangular matrix with (i, j)th entry uij. Let B 5 U 21 with (i, j)th entry bij. Let I 5 (eij ) be the identity matrix. Since U is upper triangular, uij 5 0 for all i. j. Now I 5 BU; therefore, X 1 5 e11 5 b1k uk1 5 b11 u11 k since u21 5 ? ? ? 5 un1 5 0. Therefore, u11 Þ 0 and b11 5 16 u11. For h. 1, X 0 5 eh1 5 bhk uk1 5 bh1 u11. k Since u11 Þ 0, bh1 5 0 for h. 1. Now, work with column 2 of B. X 1 5 e22 5 b2k uk2 5 b21 u12 1 b22 u22 5 b22 u22 k since b21 5 0. Therefore, u22 Þ 0 and b22 5 16 u22. For h. 2, X 0 5 eh2 5 bhk uk2 5 bh1 u12 1 bh2 u22 5 0 1 bh2 u22. k Since u22 Þ 0, bh2 5 0. We conclude that bh2 5 0 for all h. 2. This argument shows bhj 5 0 for all h. j; that is, B too is upper triangular. The second part follows by transposing the first part and Theorem 8.10b.  ANSWERS PAMPHLET 35 8.31 The (i, j)th entry of P T P is the product of the ith row of P T and the jth column of P, that is, the product of the ith column of P and the jth column of P. This product is 0 if i Þ j and 1 if i 5 j; that is, P T P 5 I. 8.32 The criterion for invertability is a11 a22 a33 2 a11 a23 a32 1 a13 a21 a31 2 a11 a21 a33 1 a11 a22 a31 2 a12 a21 a31 Þ 0. See Section 26.1. 8.33 a) Suppose a k 3 l matrix A has a left inverse L (which must be k 3 k) and a right inverse R (which must be l 3 l). Then LAR 5 (LA)R 5 IR 5 R and LAR 5 L(AR) 5 LI 5 L, so R 5 L. This is impossible since the two matrices are of different sizes. b, c) Suppose A is m 3 n with m , n. If A has rank m, then Ax 5 b has infinitely many solutions for every right-hand side b, by Fact 7.11a. Let ei 5 (0,... , 0, 1, 0,... , 0) with a 1 in the ith entry. Let ci be one of the (infinitely many) solutions of Ax 5 ei. Then, A ? [c1 ? ? ? cm ] 5 [e1 ? ? ? em ] 5 I. So, C 5 [c1 c2 ? ? ? cm ] is one of the right inverses of A. Conversely, if A has a right inverse C, then the solution of Ax 5 b is x 5 Cb since A(Cb) 5 (AC)b 5 b. By Fact 7.7, A must have rank m 5 number of rows of A. d) If A is m 3 n with m. n, apply the previous analysis to AT.         1 14 2 20 8.34 a) (I 2 A)  1  5  8 ; 21 b) (I 2 A) 21  1  5  14 ; 1 8 1 14     2 18 c) (I 2 A)21  1  5  16 . 2 18 µ ¶ a b 8.35 For with a, b, c, d. 0, a 1 c , 1, and b 1 d , 1, c d µ ¶21 µ ¶ 1 2 a 2b 1 12d b (I2A)21 5 5. 2c 1 2 d (1 2 a)(1 2 d) 2 bc c 12a Since a 1 c , 1, c , (1 2 a); since b 1 d , 1, b , (1 2 d). Therefore, 0 , bc , (1 2 a)(1 2 d) and (1 2 a)(1 2 d) 2 bc. 0. So, (I 2 A)21 is a positive matrix. 36 MATHEMATICS FOR ECONOMISTS 8.36 Let a?1 5 # of columns of A11 5 # of columns of A21. Let a?2 5 # of columns of A12 5 # of columns of A22. Let c1? 5 # of rows of C11 5 # of rows of C12 5 # of rows of C13. Let c2? 5 # of rows of C21 5 # of rows of C22 5 # of rows of C23. Then, a?1 5 c1? and a?2 5 c2?.   C11 C12 C13 8.37 C should be written as  C21 C22 C23 . In the notation of the previous C31 C32 C33 problem a?1 5 c1? 5 2 a?2 5 c2? 5 1 a?3 5 c3? 5 3.   A21 11 0 ??? 0  0 A21 ??? 0   22  8.38 A21 5....... . .....  0 0 ? ? ? A21 nn 8.39 In the notation of Exercise 8.36, A11 is of size a1? 3 a?1 ; A12 A21 22 A21 is also of size a1? 3 a?1 , so D is well defined. µ ¶Ã ! A11 A12 D21 2D21 A12 ? A21 22 A21 A22 2A21 22 A21 D 21 A21 (I 1 A D21 A A21 ) 22 21 12 22 à ! A11 D21 2 A12 A21 A 22 21 D 21 2A D21 A A21 1 A A21 1 A A21 A D21 A A21 11 12 22 12 22 12 22 21 12 22 5. A21 D21 2 A22 A21 22 A21 D 21 2A21 D21 A12 A21 21 21 21 21 22 1 A22 A22 1 A22 A22 A21 D A12 A22 Write (1, 1) as (A11 2 A12 A21 22 A21 )D 21 5 DD21 5 I. Write (1, 2) as 2(A11 2 A12 A21 21 21 21 21 21 22 A21 )D A12 A22 1 A12 A22 5 2DD A12 A22 1 A12 A22 1 5 2A12 A21 21 22 1 A12 A22 5 0. Write (2, 1) as (A21 D21 2 IA21 D21 ) 5 0. Write (2, 2) as 2A21 D21 A21 A21 21 21 22 1 I 1 A21 D A12 A22 5 I. µ ¶ I 0 So the product is the identity matrix. 0 I  ANSWERS PAMPHLET 37 µ ¶ A21 21 21 21 11 (I 1 A12 C A21 A11 ) 2A11 A12 C 21 8.40 A21 5 , 2C 21 A21 A21 11 C 21 where C 5 A22 2 A21 A21 11 A12. 8.41 a) A11 and A22 nonsingular. b) A11 and A11 2 (16 a22 )A12 A21 nonsingular. c) A22 invertible and pT A21 22 p nonzero. 8.42 a) E12 (3). b) E12 (23), E13 (2), E23 (21). c) E12 (22), E13 (3), E23 (1). d) E12 (23), E14 (22), E23 (1), E34 (21). µ ¶µ ¶ 1 0 2 4 8.43 a) , 23 1 0 21    1 0 0 2 1 0 b)  3 1 0   0 21 6 , 22 1 1 0 0 3    1 0 0 2 4 0 1 c)  2 1 0   0 22 3 1 , 23 21 1 0 0 3 8    1 0 0 0 2 6 0 5 3  1 0 0 0 3  8 2. d)  0 21 1 0   0 0 24 9 2 0 1 1 0 0 0 26 8.44 Suppose we can write A as A 5 L1 U1 5 L2 U2 where L1 and L2 are lower triangular with only 1s on the diagonals. The proof and statement of Exercise 8.30 show that L1 and L2 are invertible and that L121 and L221 are lower triangular. Since U1 5 L121 A and U2 5 L221 A, U1 and U2 are invertible too. Write L1 U1 5 L2 U2 as L221 L1 5 U2 U121. By Exercise 8.8, L221 L1 is lower triangular and U2 U121 is upper triangular. Therefore, L221 L1 and U2 U121 are both diagonal matrices. Since L1 and L2 have only 1s on the diagonal, L221 and L221 L1 have only 1s on the diagonal. It follows that L221 L1 5 I and that U2 U121 5 I. Therefore, L2 5 L1 and U2 5 U1 , and the LU decomposition of A is unique. 8.45 Suppose A 5 L1 U1 5 L2 U2 , as in the last exercise. First, choose U2 to be a row echelon matrix of A. By rearranging the order of the variables, we can assume that each row of U2 has exactly one more leading zero than the 38 MATHEMATICS FOR ECONOMISTS previous row and that its U11 Þ 0. Write L1 U1 5 L2 U2 as LU1 5 U2 where L is the lower-triangular matrix L221 L1 and has only 1s on its diagonal. Let L 5 ((lij )), U1 5 ((vij )), and U2 5 ((uij )). X 0 Þ u11 5 l1j v j1 5 l11 v11 5 v11 =⇒ v11 Þ 0. j P For k. 1, uk1 5 vk1 5 0. So, 0 5 uk1 5 nj51 lkj v j1 5 lk1 v11 =⇒ lk1 5 0 for k. 1.   1 0   This shows that the first column of L is .. . . 0 .. . 0   Similar analysis shows that the jth column of L is    1 . We need only show 0