Verify that λ; is an eigenvalue of A and that x; is a corresponding eigenvector. A₁ = 5, x₁ = (1, 2, -1) A₂ = -3, x₂ = (-2, 10) A3 = -3, x3 = (3, 0, 1) Ax1 = Ax2= Ax3 = A = -2 -2 -1 -2 -2 -1 -2 2 -3 2 1 -6 -2 2 -3 2 1 -6 0 -1 -2 2 = DE 2 -3 2 1 -6 0 -1 -2 IJ 0 2-3 3 I 2 1 -6 0 = E 5 = -3 N7 2 -1- 3 -3 0 = 2₁x1₁ = 1₂x2 = 13x3
Verify that λ; is an eigenvalue of A and that x; is a corresponding eigenvector. A₁ = 5, x₁ = (1, 2, -1) A₂ = -3, x₂ = (-2, 10) A3 = -3, x3 = (3, 0, 1) Ax1 = Ax2= Ax3 = A = -2 -2 -1 -2 -2 -1 -2 2 -3 2 1 -6 -2 2 -3 2 1 -6 0 -1 -2 2 = DE 2 -3 2 1 -6 0 -1 -2 IJ 0 2-3 3 I 2 1 -6 0 = E 5 = -3 N7 2 -1- 3 -3 0 = 2₁x1₁ = 1₂x2 = 13x3
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter7: Eigenvalues And Eigenvectors
Section7.1: Eigenvalues And Eigenvectors
Problem 66E: Show that A=[0110] has no real eigenvalues.
Related questions
Question
![1. [-/5 Points]
AX1
Verify that ; is an eigenvalue of A and that x; is a corresponding eigenvector.
2 -3 A₁ = 5, x₁ = (1, 2, -1)
λ₂ = -3, x₂ = (-2, 10)
023-3, x3 = (3, 0, 1)
1
-6
-1 -2
-
Ax 2 =
A =
Ax 3 =
-2
2
-1
-2
DETAILS
2
-2
-1
-2
2
-1 -2
2 -3
1 -6
-2
1
I
2 -3
1 -6
0
0
-2
2 -3
1-6
N
T
13
0
I
LARLINALG8 7.1.004.
=
-DE-D-
11
L
2
-1
= -3
-2
[]
= 2₁x1
3
-3 0
= 1₂x₂
= 13x3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a5e350f-0835-49bf-9d6a-aba123106e53%2F24305b8b-299a-49f7-b1d7-10292e776971%2Ff7896te_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. [-/5 Points]
AX1
Verify that ; is an eigenvalue of A and that x; is a corresponding eigenvector.
2 -3 A₁ = 5, x₁ = (1, 2, -1)
λ₂ = -3, x₂ = (-2, 10)
023-3, x3 = (3, 0, 1)
1
-6
-1 -2
-
Ax 2 =
A =
Ax 3 =
-2
2
-1
-2
DETAILS
2
-2
-1
-2
2
-1 -2
2 -3
1 -6
-2
1
I
2 -3
1 -6
0
0
-2
2 -3
1-6
N
T
13
0
I
LARLINALG8 7.1.004.
=
-DE-D-
11
L
2
-1
= -3
-2
[]
= 2₁x1
3
-3 0
= 1₂x₂
= 13x3
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