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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

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter8: Introduction To Functions
Section8.8: Linear And Quadratic Functions
Problem 4ST
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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
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 = ...
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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