Explanation Given: The given system is, d x d t = 3 x − y − z d y d t = x + y − z d z d t = x − y + z Calculation: For the given system, the coefficient matrix is A = ( 3 − 1 − 1 1 1 − 1 1 − 1 1 ) . The characteristic equation for the matrix A is | ( A − λ I ) | = 0 where, I is the identity matrix. Substitute the values of matrix A and I in above expression. | 3 − λ − 1 − 1 1 1 − λ − 1 1 − 1 1 − λ | = 0 ( 3 − λ ) [ ( 1 − λ ) 2 − 1 ] + 1 [ 1 − λ + 1 ] − 1 [ − 1 + λ − 1 ] = 0 ( 3 − λ ) [ λ 2 − 2 λ ] + ( 2 − λ ) + ( 2 − λ ) = 0 ( 3 − λ ) ( λ − 2 ) λ + ( 2 − λ ) + ( 2 − λ ) = 0 Simplify further, ( λ − 2 ) [ ( 3 − λ ) λ − 1 − 1 ] = 0 ( λ − 2 ) ( 3 λ − λ 2 − 2 ) = 0 ( λ − 2 ) ( λ − 2 ) ( 1 − λ ) = 0 ( 1 − λ ) ( λ − 2 ) 2 = 0 Therefore, the eigenvalues of the above system is λ 1 = 1 with multiplicity 1 and λ 2 = 2 with multiplicity 2 . Obtain the eigenvector K 1 corresponding to the eigenvalue, λ 1 = 1 . Solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 z 1 ) . Substitute the values of A , λ 1 and K 1 in the above equation gives, [ ( 3 − 1 − 1 1 1 − 1 1 − 1 1 ) − ( 1 0 0 0 1 0 0 0 1 ) ] ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 2 − 1 − 1 1 0 − 1 1 − 1 0 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) By applying the row transformation R 3 → R 3 − R 2 in the above system of equations, ( 2 − 1 − 1 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Now, by row transformation R 1 → R 1 − 2 R 2 . ( 0 − 1 1 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) Which on applying R 1 → R 1 − R 3 in above system, ( 0 0 0 1 0 − 1 0 − 1 1 ) ⋅ ( x 1 y 1 z 1 ) = ( 0 0 0 ) ( 0 x 1 − z 1 − y 1 + z 1 ) = ( 0 0 0 ) Solve the above system of equation gives, x 1 = z 1 and y 1 = z 1 Assume that z 1 = p 1 for some scalar p 1 , which implies, x 1 = p 1 , y 1 = p 1 Thus, the value of ( x 1 y 1 z 1 ) is, ( x 1 y 1 z 1 ) = ( p 1 p 1 p 1 ) = p 1 ( 1 1 1 ) Therefore, the eigenvector K 1 corresponding to λ 1 = 1 is ( 1 1 1 )

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

11th Edition

ISBN: 9781305965720

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 8.2, Problem 25E

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To find: The general solution of the given system.

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(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).

Let us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…

= Consider the hypothesis test Ho: μ₁ = μ₂ against H₁ μ₁ μ2. Suppose that sample sizes are n₁ = 15 and n₂ = 15, that x1 = 4.7 and X2 = 7.8 and that s² = 4 and s² = 6.26. Assume that o and that the data are drawn from normal distributions. Use απ 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain ẞ = 0.05 if the true difference in means is - 2? Assume that α = 0.05. (a) The null hypothesis is 98.7654). rejected. The P-value is 0.0008 (b) The power is 0.94 . Round your answer to four decimal places (e.g. Round your answer to two decimal places (e.g. 98.76). (c) n₁ = n2 = 1 . Round your answer to the nearest integer.

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Solution Summary: The author explains that the general solution of the given system is X(t)=c_1 left, where I is the identity matrix.

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