0.48 - 0.02 0.38 0.2 Let A= 0.15 0.85 0.15 The vector v, = 0.5 is an eigenvector for A, and two eigenvalues are 0.7 and 0.1. Construct the solution of the dynamical system x,1= Ax, that satisfies x, = (0.3, - 0.1,0.8). What happens to x, as k+o? 0.37 0.17 0.47 0.3 Find the solution of the equation x +1 = Ax. Choose the correct answer below. 0.2 2 -1 O A. X = 0.5 + (0.7)k + (0.1)k -3 0.3 1 -1 0.2 O B. Xx= -3 +0.2(0.7)k 0 +0.3(0.1)k 0.5 0.3 0.2 2 Oc. X = 0.5 +0.2(0.7)k - 3 +0.3(0.1)k 0.3 1 Describe what happens to x, as ko. Choose the correct answer below. O A. As k becomes sufficiently large, x 0.2(0.7)* - 3 + 0.3(0.1)k 1 -1 O B. As k becomes sufficiently large, x, 0.3(0.1)* 0.2 Oc. As k becomes sufficiently large, x, 0.5 0.3
0.48 - 0.02 0.38 0.2 Let A= 0.15 0.85 0.15 The vector v, = 0.5 is an eigenvector for A, and two eigenvalues are 0.7 and 0.1. Construct the solution of the dynamical system x,1= Ax, that satisfies x, = (0.3, - 0.1,0.8). What happens to x, as k+o? 0.37 0.17 0.47 0.3 Find the solution of the equation x +1 = Ax. Choose the correct answer below. 0.2 2 -1 O A. X = 0.5 + (0.7)k + (0.1)k -3 0.3 1 -1 0.2 O B. Xx= -3 +0.2(0.7)k 0 +0.3(0.1)k 0.5 0.3 0.2 2 Oc. X = 0.5 +0.2(0.7)k - 3 +0.3(0.1)k 0.3 1 Describe what happens to x, as ko. Choose the correct answer below. O A. As k becomes sufficiently large, x 0.2(0.7)* - 3 + 0.3(0.1)k 1 -1 O B. As k becomes sufficiently large, x, 0.3(0.1)* 0.2 Oc. As k becomes sufficiently large, x, 0.5 0.3
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter4: Eigenvalues And Eigenvectors
Section4.6: Applications And The Perron-frobenius Theorem
Problem 69EQ: Let x=x(t) be a twice-differentiable function and consider the second order differential equation...
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