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= A = 4 -1 4 0 2 1 0 03 4 -1 4 (3) |_| 0 2 1 0 0 03 0 I 4 -1 4 A₁ = 4, x₁ = (1, 0, 0) A₂ = 2, x₂ = (1, 2, 0) 13 = 3, x3 = (-3, 1, 1) 4 -1 4 1 -(6 393) - ]] 0 21 = = 0 03 2 1 03 = -48 = -3 1 = -0-||--0- 1 1 = 2 = - - - 0 = 3 : 21x1 1 = 1 : 12x2 = 13x3

= A = 4 -1 4 0 2 1 0 03 4 -1 4 (3) |_| 0 2 1 0 0 03 0 I 4 -1 4 A₁ = 4, x₁ = (1, 0, 0) A₂ = 2, x₂ = (1, 2, 0) 13 = 3, x3 = (-3, 1, 1) 4 -1 4 1 -(6 393) - ]] 0 21 = = 0 03 2 1 03 = -48 = -3 1 = -0-||--0- 1 1 = 2 = - - - 0 = 3 : 21x1 1 = 1 : 12x2 = 13x3

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Verify that ; is an eigenvalue of A and that x; is a corresponding eigenvector. A₁ = 4, X₁ = (1, 0, 0) A2 = 2, x₂ = (1, 2, 0) A3 = 3, x3 = (-3, 1, 1) Ax1 = ~¤¤·|····~ Ax2 = Ax3 A = = 4 -1 4 0 21 0 0 3 = 4-1 4 4 -1 4 ---|| 2 1 = 03 0 21 0 0 0 3 0 I 2 1 03 ↓ 1 = 4 = 2 0 = 3 0 12 4 -1 4 -3 1 = -||--0- 1 ↓ 1 O -3 = 1 = 12x2 = 13×3
Transcribed Image Text:Verify that ; is an eigenvalue of A and that x; is a corresponding eigenvector. A₁ = 4, X₁ = (1, 0, 0) A2 = 2, x₂ = (1, 2, 0) A3 = 3, x3 = (-3, 1, 1) Ax1 = ~¤¤·|····~ Ax2 = Ax3 A = = 4 -1 4 0 21 0 0 3 = 4-1 4 4 -1 4 ---|| 2 1 = 03 0 21 0 0 0 3 0 I 2 1 03 ↓ 1 = 4 = 2 0 = 3 0 12 4 -1 4 -3 1 = -||--0- 1 ↓ 1 O -3 = 1 = 12x2 = 13×3
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