Program Explanation Given information: The system is x ′ = [ 2 1 0 1 0 2 1 0 0 0 2 1 0 0 0 2 ] x . Explanation: Consider matrix form of the system of equation as A . A = [ 2 1 0 1 0 2 1 0 0 0 2 1 0 0 0 2 ] The characteristics equation of the coefficient matrix is, [ 2 − λ 1 0 1 0 2 − λ 1 0 0 0 2 − λ 1 0 0 0 2 − λ ] = 0 ( λ − 2 ) 4 = 0 λ = 2 , 2 , 2 and 2 The obtain Eigen value has multiplicity 4. The Eigen vector of the coefficient matrix A is in the form as, [ 2 − λ 1 0 1 0 2 − λ 1 0 0 0 2 − λ 1 0 0 0 2 − λ ] [ a b c d ] = [ 0 0 0 0 ] The associated Eigen vector with the matrix A is v = [ a b c d ] T . Substitute 1 for λ in coefficient matrix, [ 2 − λ 1 0 1 0 2 − λ 1 0 0 0 2 − λ 1 0 0 0 2 − λ ] [ a b c d ] = [ 0 0 0 0 ] [ 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 ] [ a b c d ] = [ 0 0 0 0 ] The scalar equation are shown below, b + d = 0 c = 0 d = 0 b = 0 Therefore, the associated Eigen vector is obtained as: u 1 = [ 1 0 0 0 ] The other three associated vectors can be obtained by the given formula, v 3 = ( A − 2 I ) u v 2 = ( A − 2 I ) v 3 v 1 = ( A − 2 I ) v 2 The value of ( A − 2 I ) 2 can be obtained as, ( A − 2 I ) 2 = [ 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 ] [ 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 ] = [ 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 ] The value of ( A − 2 I ) 3 can be obtained as, ( A − 2 I ) 3 = [ 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 ] [ 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 ] = [ 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ] The value of ( A − 2 I ) 4 can be obtained as, ( A − 2 I ) 4 = [ 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ] [ 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ] = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ] Therefore, any vector u ≠ 0 is a solution of the equation ( A − 2 I ) 4 u = 0 . Consider the vector u = [ 0 0 0 1 ] The associated vector u = v 4

5th Edition

ISBN: 9780135904206

Author: Edwards

Publisher: PEARSON

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Problems on Growing Functions

For a given set of functions f and g, the growth of functions states that the function g grows as fast as the function f. The growth rate of algorithms helps determine the algorithm's efficiency and compare its performance to other algorithms.

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Chapter 5.5, Problem 20P

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[2101021000210002]x .

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Chapter 5.5, Problem 19P

Chapter 5.5, Problem 21P

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

Chapter 5 Solutions

DIFFERENTIAL EQUATIONS+..-MYLAB ACCESS

Ch. 5.1 - Let A=[2347] and B=[3451]. Find (a) 2A+3B; (b)...Ch. 5.1 - Prob. 2P Ch. 5.1 - Find AB and BA given A=[203415] and B=[137032].Ch. 5.1 - Prob. 4P Ch. 5.1 - Prob. 5P Ch. 5.1 - Prob. 6P Ch. 5.1 - Prob. 7P Ch. 5.1 - Prob. 8P Ch. 5.1 - Prob. 9P Ch. 5.1 - Prob. 10P

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Program Description: Purpose of problem is to obtain the general solution of the system x ′ = [ 2 1 0 1 0 2 1 0 0 0 2 1 0 0 0 2 ] x .

Solution Summary: The author explains that the problem is to obtain the general solution of the system of equation as A.

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Program Description:Purpose of problem is to obtain the general solution of the system x′=[2101021000210002]x .

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Chapter 5 Solutions