28 a w t 1 0 8 w t 0 1 m t m t 1 0 2 w t 0 9 m t so a

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Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 9 / Exercise 26
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28. a. w ( t + 1) = 0 . 8 w ( t ) + 0 . 1 m ( t ) m ( t + 1) = 0 . 2 w ( t ) + 0 . 9 m ( t ) so A = 0 . 8 0 . 1 0 . 2 0 . 9 which is a regular transition matrix since its columns sum to 1 and its entries are positive. b. The eigenvectors of A are 0 . 1 0 . 2 or 1 2 with λ 1 = 1, and 1 - 1 with λ 2 = 0 . 7. x 0 = 1200 0 = 400 1 2 + 800 1 - 1 so x ( t ) = 400 1 2 + 800(0 . 7) t 1 - 1 or w ( t ) = 400 + 800(0 . 7) t m ( t ) = 800 - 800(0 . 7) t . c. As t → ∞ , w ( t ) 400 so Wipfs won’t have to close the store. 29. The i th entry of Ae is [ a i 1 a i 2 · · · a in ] e = n j =1 a ij = 1, so Ae = e and λ = 1 is an eigenvalue of A , corresponding to the eigenvector e . 344
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Applied Calculus for the Managerial, Life, and Social Sciences
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Chapter 9 / Exercise 26
Applied Calculus for the Managerial, Life, and Social Sciences
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ISM: Linear Algebra Section 7.2 30. a. Let v i be the largest component of the vector v , that is, v i v j for j = 1 , . . . , n . Then the i th component of A v is λ v i = n j =1 a ij v j n j =1 a ij v i = n j =1 a ij v i = v i v j v i n j =1 a ij = 1 We can conclude that λ v i v i , and therefore λ 1, as claimed. Also note that if v is not a multiple of the eigenvector e discussed in Exercise 29, then v j < v i for some index j , so that n j =1 a ij v j < n j =1 a ij v i and therefore λ < 1. b. Let v i be the component of v with the largest absolute value, that is, | v i | | v j | for j = 1 , 2 , . . . , n . Then the absolute value of the i th component of Av is | λ v i | = n j =1 a ij v j n j =1 a ij | v j | n j =1 a ij | v i | = n j =1 a ij | v i | = | v i | so that | λ v i | | v i | and | λ | 1, as claimed. 31. Since A and A T have the same eigenvalues (by Exercise 22), Exercise 29 states that λ = 1 is an eigenvalue of A , and Exercise 30 says that | λ | 1 for all eigenvalues λ . Vector e need not be an eigenvector of A ; consider A = 0 . 9 0 . 9 0 . 1 0 . 1 . 32. f A ( λ ) = - λ 3 +3 λ + k . The eigenvalues of A are the solutions of the equation - λ 3 +3 λ + k = 0 , or, λ 3 - 3 λ = k . Following the hint, we graph the function g ( λ ) = λ 3 - 3 λ as shown in Figure 7.14. We use the derivative f ( λ ) = 3 λ 2 - 3 to see that g ( λ ) has a global minimum at (1 , - 2) and a global maximum at ( - 1 , 2). To count the eigenvalues of A , we need to find out how many times the horizontal line y = k intersects the graph of g ( λ ). In Figure 7.14, we see that there are three solutions if k satisfies the inequality 2 > k > - 2, two solutions if k = 2 or k = - 2, and one solution if | k | > 2. 33. a. f A ( λ ) = det( A - λ I 3 ) = - λ 3 + c λ 2 + b λ + a 345
Chapter 7 ISM: Linear Algebra ( - 1, 2) g ( λ ) = λ 3 - 3 λ (1, - 2) Figure 7.14: for Problem 7.2.32. b. By part a, we have c = 17, b = - 5 and a = π , so M = 0 1 0 0 0 1 π - 5 17 . 34. Consider the possible graphs of f A ( λ ) assuming that it has 2 distinct real roots. ( 1 λ ) (– λ – 2) ( λ 2 + 1) Figure 7.15: for Problem 7.2.34. Algebraic multiplicity of each eigenvalue is 1. Example: 1 0 0 0 0 - 2 0 0 0 0 0 - 1 0 0 1 0 . See Figure 7.15. Algebraic multiplicity of each eigenvalue is 2. Example: - 2 0 0 0 0 - 2 0 0 0 0 1 0 0 0 0 1 . See 346
ISM: Linear Algebra Section 7.2 (– λ – 2) 2 (1 λ ) 2 Figure 7.16: for Problem 7.2.34. Figure 7.16. (– λ – 2) (1 λ ) 3 Figure 7.17: for Problem 7.2.34.

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