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Iterative Methods-Numerical Analysis-Quiz Solution

Exams, Numerical Analysis

Post: July 31th, 2012
Description
This is solution to quiz in class of Numerical Analysis. It was taken by Prof. Anuha Sharma at Biyani Girls College. It includes: AIterative, Methods, Numerical, Analysis, System, Linear, Equations, Infinity, Norm, Solution
This is solution to quiz in class of Numerical Analysis. It was taken by Prof. Anuha Sharma at Biyani Girls College. It includes: AIterative, Methods, Numerical, Analysis, System, Linear, Equations, Infinity, Norm, Solution
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Contents
1 Quiz 2, Solution Question 1 1 1 1 x2 x3 + 3 3 3 1 1 = 0; x2 = x1 x3 2 3 43 3 = 4; x3 = x2 x1 77 7 = 1; x1 = (0) (0) t 3x1 x2 + x3 3x1 + 6x2 + 2x3 3x1 + 3x2 + 7x3 (0) x(0) = x1 ; x2 ; x3 = (0; 0; 0) t 1 x1 = 3 x2 1 x3 + 1 x =0;x =0 = x1 = 0:333 33 3 32 3 x2 = 1 x1 1 x3 x1 =0:333 33;x3 =0 = x2 = 0:166 67 2 3 x3 = 4 3 x2 3 x1 x1 =0:333 33;x2 = 0:166 67 = x3 = 0:5 7 7 7 x(1) = x1 ; x2 ; x3 (1) (1) (1) t = (0:333 33; 0:166 67; 0:5) t x1 = 1 x2 1 x3 + 1 x2 = 0:166 67;x3 =0:5 = x1 = 0:111 11 3 3 3 1 x2 = 2 x1 1 x3 x =0:111 11;x =0:5 = x2 = 0:222 22 3 1 3 4 x3 = 7 3 x2 3 x1 x1 =0:111 11;x2 = 0:222 22 = x3 = 0:619 05 7 7 x(2) = x1 ; x2 ; x3 (2) (2) (2) t = (0:111 11; 0:222 22; 0:61905) t max (j0:111 11 x(2) x(1) 1 0:333 33j ; j 0:222 22 + 0:166 67j ; j0:619 05 0:5j) = 0:222 22 x(2) 1 = max (j0:11111 0:33333j ; j 0:22222 + 0:16667j ; j0:61905 max (0:52381; 0:222 22; 0:52381) 0:5j) = 0:424 24 1 docsity.com Question 2 1 9 x2 + 10 10 1 1 7 7; x2 = x1 + x3 + 10 5 1..

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