After three iterations, the Jacobi method applied with initial approximation (0) .(0) .(0) = x3 = 0.1 to the system X1 + 3.x2 – x3 | 3.x1 – x2 = 5 x2 + 2.x3 = 1
After three iterations, the Jacobi method applied with initial approximation (0) .(0) .(0) = x3 = 0.1 to the system X1 + 3.x2 – x3 | 3.x1 – x2 = 5 x2 + 2.x3 = 1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
What is x1, x2 and x3?
![After three
iterations, the
Jacobi method
applied with
initial
approximation
(0)
= X2' =
.(0)
.(0)
= x3
= 0.1
to the system
x1 + 3.x2 – x3
3x1 – x2 = 5
x2 + 2x3 = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d23c93b-870e-4ad1-8eae-60e3d2381ea8%2F623a6dca-e423-40ed-b0c0-b92cfd75270c%2Fj2p5a8l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:After three
iterations, the
Jacobi method
applied with
initial
approximation
(0)
= X2' =
.(0)
.(0)
= x3
= 0.1
to the system
x1 + 3.x2 – x3
3x1 – x2 = 5
x2 + 2x3 = 1
![.(0)
.(0)
= 12
.(0)
X3
= 0.1
to the system
xị + 3.x2 – x3
|
3.x1 – x2 = 5
X2 + 2.x3 =
1
(after it has been
re-arranged so that
it is strictly
diagonally
dominant), gives
the solution of the
system as
X1 = ..., x2
... and x3 =
...](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d23c93b-870e-4ad1-8eae-60e3d2381ea8%2F623a6dca-e423-40ed-b0c0-b92cfd75270c%2Fcboa8ve_processed.jpeg&w=3840&q=75)
Transcribed Image Text:.(0)
.(0)
= 12
.(0)
X3
= 0.1
to the system
xị + 3.x2 – x3
|
3.x1 – x2 = 5
X2 + 2.x3 =
1
(after it has been
re-arranged so that
it is strictly
diagonally
dominant), gives
the solution of the
system as
X1 = ..., x2
... and x3 =
...
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