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Verify that λ, is an eigenvalue of A and that x; is a corresponding eigenvector. -4-2 A = -2 -7 A₁ =-11, x₁ = (1, 2, -1) λ₂ = -3, x₂ = (-2, 10) 23 = -3, x3 = (3, 0, 1) 1 _Ax₁ = -4 -2 -2 -7 1 6 --- 2-6 ↓1 -4 -2 -2 -7 1 3 6 _Ax₂ = = [1]] =2₂*2 2-6 -4 -2 3 ---- _AX3 = -2 -7 6 = = √3x3 1 2-6 36 2-6 2 = ₁₁

Verify that λ, is an eigenvalue of A and that x; is a corresponding eigenvector. -4-2 A = -2 -7 A₁ =-11, x₁ = (1, 2, -1) λ₂ = -3, x₂ = (-2, 10) 23 = -3, x3 = (3, 0, 1) 1 _Ax₁ = -4 -2 -2 -7 1 6 --- 2-6 ↓1 -4 -2 -2 -7 1 3 6 _Ax₂ = = [1]] =2₂*2 2-6 -4 -2 3 ---- _AX3 = -2 -7 6 = = √3x3 1 2-6 36 2-6 2 = ₁₁

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Verify that λ; is an eigenvalue of A and that x; is a corresponding eigenvector. -4 -2 A = -2 -7 6 3 A₁ =-11, x₁ = (1, 2, -1) λ₂ = -3, x₂ = (-2, 10) 2-6 A3 = -3, x3 = (3, 0, 1) 1 -4 -2 3 1 Ax₁ = -2 -7 6 2 = 2 = ₁₁ 1 2 -6 DE+ DE -4 -2 3 -2 -2 ~H AX2 = -2 -7 6 1 = = 1 = 2₂x2 1 2 -6 0 0 -4 -2 3 3 3 -DE- AX3 = -2 -7 6 0 = = 1 2 -6 1 a w → -3 -3 = 13x3
Transcribed Image Text:Verify that λ; is an eigenvalue of A and that x; is a corresponding eigenvector. -4 -2 A = -2 -7 6 3 A₁ =-11, x₁ = (1, 2, -1) λ₂ = -3, x₂ = (-2, 10) 2-6 A3 = -3, x3 = (3, 0, 1) 1 -4 -2 3 1 Ax₁ = -2 -7 6 2 = 2 = ₁₁ 1 2 -6 DE+ DE -4 -2 3 -2 -2 ~H AX2 = -2 -7 6 1 = = 1 = 2₂x2 1 2 -6 0 0 -4 -2 3 3 3 -DE- AX3 = -2 -7 6 0 = = 1 2 -6 1 a w → -3 -3 = 13x3
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