Program Explanation Given information: The differential equation is x ′ 1 = 5 x 1 − 6 x 3 , x ′ 2 = 2 x 1 − x 2 − 2 x 3 and x ′ 3 = 4 x 1 − 2 x 2 − 4 x 3 . Explanation: Consider matrix form of the system of equation as A. A x = [ 5 0 − 6 2 − 1 − 2 4 − 2 − 4 ] x The characteristics equation of the coefficient matrix is, [ 5 − λ 0 − 6 2 − 1 − λ − 2 4 − 2 − 4 − λ ] = 0 ( 5 − λ ) [ − 1 − λ 1 1 − 4 − λ ] − 0 [ 2 − 2 4 4 − λ ] − 6 [ 2 − 1 − λ 4 − 2 ] = 0 − λ 3 + λ = 0 λ ( 1 − λ 2 ) = 0 Therefore, the Eigen values of the matrix can be obtained as, λ ( 1 − λ 2 ) = 0 λ = 1 or λ = 1 or λ = 0 The Eigen vector of the coefficient matrix A is in the form as, [ 5 − λ 0 − 6 2 − 1 − λ − 2 4 − 2 − 4 − λ ] [ a b c ] = [ 0 0 0 ] The associated Eigen vector with the matrix Ais v = [ a b c ] T . Substitute 9 for λ in coefficient matrix, [ 5 − 0 0 − 6 2 − 1 − 0 − 2 4 − 2 − 4 − 0 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] [ 5 0 − 6 2 − 1 − 2 4 − 2 − 4 ] [ a 1 b 1 c 1 ] = [ 0 0 0 ] The scalar equation are shown below, 5 a 1 − 6 c 1 = 0 2 a 1 − b 1 − 2 c 1 = 0 4 a 1 − 2 b 1 − 4 c 1 = 0 After solving the equation the value of a 1 , b 1 and c 1 can be obtained as, a 1 = 6 b 1 = 2 c 1 = 5 Therefore, the associated Eigen vector is shown below. v 1 = [ 6 2 5 ] Substitute 1 for λ in coefficient matrix, [ 5 − 1 0 − 6 2 − 1 − 1 − 2 4 − 2 − 4 − 1 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] [ 4 0 − 6 2 − 2 − 2 4 − 2 − 5 ] [ a 2 b 2 c 2 ] = [ 0 0 0 ] The scalar equation are shown below, 4 a 2 − 6 c 2 = 0 2 a 2 − 2 b 2 − 2 c 2 = 0 4 a 2 − 2 b 2 − 5 c 2 = 0 After solving the equation the value of a 2 , b 2 and c 2 can be obtained as, a 2 = 3 b 2 = 1 c 2 = 2 Therefore, the associated Eigen vector is shown below

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ISBN: 9780100802452

Author: Calvis

Publisher: YUZU

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Chapter 5.2, Problem 21P

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1−6x3,x′2=2x1−x2−2x3 and x′3=4x1−2x2−4x3 .

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Chapter 5.2, Problem 20P

Chapter 5.2, Problem 22P

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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.

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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 x ′ 1 = 5 x 1 − 6 x 3 , x ′ 2 = 2 x 1 − x 2 − 2 x 3 and x ′ 3 = 4 x 1 − 2 x 2 − 4 x 3 .

Solution Summary: The author explains that the purpose of the problem is to obtain the general solution of xprime.

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Program Description:Purpose of problem is to obtain the general solution of x′1=5x1−6x3,x′2=2x1−x2−2x3 and x′3=4x1−2x2−4x3 .

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Chapter 5 Solutions