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Verify that 2, is an eigenvalue of A and that x, is a corresponding eigenvector. = 13, x, = (1, 2, –1) h2 = -3, x, = (-2, 1 0) 23 = -3, x3 = (3, 0, 1) -1 4 -6 A = 4 5 -12 -2 -4 3 -1 4 -6 1 Ах, 5 -12 2 = 1,x1 4 2 13 -2 -4 -1 -1 4 -6 Ax2 = = -3 1 = 1,x2 4 5 -12 -2 -4 -1 4 -6 3. 3 Ax3 = -3 0 = 13x3 4 5 -12 = -2 -4 1.

Verify that 2, is an eigenvalue of A and that x, is a corresponding eigenvector. = 13, x, = (1, 2, –1) h2 = -3, x, = (-2, 1 0) 23 = -3, x3 = (3, 0, 1) -1 4 -6 A = 4 5 -12 -2 -4 3 -1 4 -6 1 Ах, 5 -12 2 = 1,x1 4 2 13 -2 -4 -1 -1 4 -6 Ax2 = = -3 1 = 1,x2 4 5 -12 -2 -4 -1 4 -6 3. 3 Ax3 = -3 0 = 13x3 4 5 -12 = -2 -4 1.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter10: Matrices
Section10.CR: Chapter 10 Review
Problem 62CR
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Verify that 2, is an eigenvalue of A and that x, is a corresponding eigenvector. = 13, x, = (1, 2, -1) 12 = -3, x2 = (-2, 1 0) з 3 -3, х3 3 (3, 0, 1) -1 4 -6 A = 4 5 -12 -2 -4 3 -1 4 -6 1. Ax1 = 2 = 1,x1 4 5 -12 = 13 -2 -4 3 -1 -1 4 -6 -2 Ax2 = 1= 1,x2 4 5 -12 1 -3 -2 -4 3 -1 4 -6 3 Ax3 = = -3 = 13X3 4 5 -12 -2 -4 3 1. 1.
Transcribed Image Text:Verify that 2, is an eigenvalue of A and that x, is a corresponding eigenvector. = 13, x, = (1, 2, -1) 12 = -3, x2 = (-2, 1 0) з 3 -3, х3 3 (3, 0, 1) -1 4 -6 A = 4 5 -12 -2 -4 3 -1 4 -6 1. Ax1 = 2 = 1,x1 4 5 -12 = 13 -2 -4 3 -1 -1 4 -6 -2 Ax2 = 1= 1,x2 4 5 -12 1 -3 -2 -4 3 -1 4 -6 3 Ax3 = = -3 = 13X3 4 5 -12 -2 -4 3 1. 1.
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