Solve the first equation for Given the matrix A, below, the system below has a nontrivial solution corresponding to the eigenvalue for matrix A. Show that this y is a complex multiple of the vector 1 X₂ in terms of x₁, and from that produce the eigenvector y = - 5+2i x₁ -5-2i 29 x₂ = which is a basis for the eigenspace corresponding to λ=0.8 -0.6i. -0.7 -0.3 8.7 2.3 (-1.5 +0.6i)x₁ - 8.7x₁ + 0.3x₂ = 0 (1.5 +0.6i)x₂ = 0 Solve the first equation, (-1.5+0.6i)x₁ -0.3x2 = 0 for x₂ in terms of X₁.
Solve the first equation for Given the matrix A, below, the system below has a nontrivial solution corresponding to the eigenvalue for matrix A. Show that this y is a complex multiple of the vector 1 X₂ in terms of x₁, and from that produce the eigenvector y = - 5+2i x₁ -5-2i 29 x₂ = which is a basis for the eigenspace corresponding to λ=0.8 -0.6i. -0.7 -0.3 8.7 2.3 (-1.5 +0.6i)x₁ - 8.7x₁ + 0.3x₂ = 0 (1.5 +0.6i)x₂ = 0 Solve the first equation, (-1.5+0.6i)x₁ -0.3x2 = 0 for x₂ in terms of X₁.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Given the matrix A, below, the system below has a nontrivial solution corresponding to the eigenvalue 0.8 -0.6i. Solve the first equation for
1
X2 in terms of x₁, and from that produce the eigenvector y =
for matrix A. Show that this y is a complex multiple of the vector
- 5+2i
×-[¯
-5-2i
29
A =
which is a basis for the eigenspace corresponding to λ=0.8 -0.6i.
-0.7 -0.3
8.7 2.3
(-1.5 +0.6i)x₁ -
8.7x₁ +
0.3x₂ = 0
(1.5 +0.6i)x₂ = 0
Solve the first equation, (-1.5 +0.6)x₁ -0.3x2 = 0 for x₂ in terms of x₁-
x₂ = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fe3cb2f-b243-4382-966f-eceede055ec9%2F9fce8336-8c22-4a40-bd2b-dc6d884c516e%2Fbbj69nh_processed.png&w=3840&q=75)
Transcribed Image Text:Given the matrix A, below, the system below has a nontrivial solution corresponding to the eigenvalue 0.8 -0.6i. Solve the first equation for
1
X2 in terms of x₁, and from that produce the eigenvector y =
for matrix A. Show that this y is a complex multiple of the vector
- 5+2i
×-[¯
-5-2i
29
A =
which is a basis for the eigenspace corresponding to λ=0.8 -0.6i.
-0.7 -0.3
8.7 2.3
(-1.5 +0.6i)x₁ -
8.7x₁ +
0.3x₂ = 0
(1.5 +0.6i)x₂ = 0
Solve the first equation, (-1.5 +0.6)x₁ -0.3x2 = 0 for x₂ in terms of x₁-
x₂ = 0
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