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Q4) (25p) Use Gauss-Seidel method to solve the following linear system. Start with (x1^(0),x2^(0),x3^(0)) = (0,0,0) and perform 2 iterations only. 3x1 + 6x2 + 2x3 = 0 3x1 + 3x2 + 7x3 = 4 3x1 - x2 + x3 = 1 

          Q4) (25p) Use Gauss-Seidel method to solve the following linear system. Start with (x1^(0),x2^(0),x3^(0)) = (0,0,0) and perform 2 iterations only.

3x1 + 6x2 + 2x3 = 0
3x1 + 3x2 + 7x3 = 4
3x1 - x2 + x3 = 1
Q4) (25p) Use Gauss-Seidel method to solve the following linear system. Start with (x1^(0),x2^(0),x3^(0)) = (0,0,0) and perform 2 iterations only. 3x1 + 6x2 + 2x3 = 0 3x1 + 3x2 + 7x3 = 4 3x1 - x2 + x3 = 1

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Q4) (25p) Use Gauss-Seidel method to solve the following linear system. Start with (x1^(0),x2^(0),x3^(0)) = (0,0,0) and perform 2 iterations only. 3x1 + 6x2 + 2x3 = 0 3x1 + 3x2 + 7x3 = 4 3x1 - x2 + x3 = 1
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00:01 Well in given question 3x1 plus 6x2 plus 2 plus 2 x3 equals to 0 and vx1 plus 3x2 equals 7 x3 equals to 4 and x x x1 minus x2 plus x3 equals to 1 1 1 the iteration of the variable x1 is given by x i of k plus 1 is equal to 1 by x x x i of k plus 1 b i minus summation of j equals to 1 to i minus 1 of a i j x j of k plus 1 and a summation of j equals to j plus 1 to n a i j of x k here and equals to 3 then x 1 k plus 1 equals to 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 minus a 1 2 x2 of k minus e b and 3 x of k 3 that equals to 1 by 3 of 0 minus 6 x2 of k minus 2 x 3 of k the f power x 1 of k plus 1 equals 2 1 by 3 of minus 6 x 2 of k minus 2 x 3 of k then x 2 equals to a by 1 by a 2 is 2 2 of p 2 minus a 2 1 to x1 of k plus 1 minus a 2 3 of k therefore x2 equals to 1 by 3 of 4 minus 3 of x 1 x1 of k plus 1 7 x3 of k then x3 equals x3 of k that equals to 1 by a of 3 3 of 3 minus a3 1 of h1 of k plus 1 minus a32 of x2 of k plus 1.
02:38 That equals to 1 by 1 of 1 minus 3 x1 of k plus 1 plus x2 of k plus 1...
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