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Solve by Gauss Jacobi Iteration Method: 5x – Y + Z = 10, 2x + 4y = 12, X + Y + 5z = -1. - Applied Mathematics 1

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Question

Solve by Gauss Jacobi Iteration Method: 5x – y + z = 10, 2x + 4y = 12, x + y + 5z = -1. 

Solution

From 1st equation, 5x = 10 + y – z

∴` x=1/5(10+y-z)=0.2(10+y-z)`

Similarly,
From `2^(nd)` equation, x + 2y = 6 

∴ 2y = 6 – x

`y=1/2(6-x)=0.5(6-x)` 

z = 0.2( - 1 – x – y) = -0.2(1 + x + y) 

Iteration 1: 

Put `x_0 = y_0 = z_0`

`∴x_1 = = 0.2(10 + y_0 – z_0) = = 0.2(10 + 0 – 0) = 2`

 `∴y_1 = 0.5(6 – x_0) = 0.5(6 – 0) = 3`

 `∴z_1 = -0.2(1 + x_0 + y_0) = -0.2(1 + 0 + 0) = -0.2`

Iteration 2:

Put` x_1 = 2;y_1 = 3; z_1 = -0.2` 

`∴y_2 = 0.5(6 – x_1) = 0.5(6 – 2) = 2`

`∴z_2 = -0.2(1 + x_1 + y_1) = -0.2(1 + 2 + 3) = -1.2`

Iteration 3: 

Put x_2 = 2.64; y_2 = 2; z_2 = -1.2
`∴x_3 = 0.2(10 + y_2 – z_2) = 0.2(10 + 2 = 1.2) = 2.64`
`∴y_3 = 0.5(6 – x_2) = 0.5(6 – 2.64) = 1.68`
`∴z_3 = -0.2(1 + x_2 + y_2) = -0.2(1 + 2.64 + 2) = -1.128`

Iteration 4:
Put `x_3 = 2.64; y_3 = 1.68; z_3 = -1.128`
`∴x4 = 0.2(10 + y_3 – z_3) = 0.2(10 + 1.68 = 1.128) = 2.5615`
`∴y_4 = 0.5(6 – x_3) = 0.5(6 – 2.64) = 1.68`
`∴z_4 = -0.2(1 + x_3 + y_3) = -0.2(1 + 2.64 + 1.68) = -1.0640`

Iteration 5:
`Put x_4 = 2.5616; y_4 = 1.68; z_4 = -1.0640`
`∴x_5 = 0.2(10 + y_4 – z_4) = 0.2(10 + 1.68 + 1.0640) = 2.5488`
`∴y_5 = 0.5(6 – x_4) = 0.5(6 – 2.5616) = 1.7172`
`∴z_5 = -0.2(1 + x_4 + y_4) = -0.2(1 + 2.5616 + 1.68) = -1.0483`

Iteration 6:
`Put x_5 = 2.5488; y_5 = 1.7192; z5 = -1.0483`
`∴x6 = 0.2(10 + y_5 – z5) = 0.2(10 + 1.7192 + 1.0483) = 2.5535`
`∴y6 = 0.5(6 – x_5) = 0.5(6 – 2.5488) = 1.7256`
`∴z_6 = -0.2(1 + x_5 + y_5) = -0.2(1 + 2.5488 + 1.7192) = -1.0536`

Iteration 7:
`Put x_6 = 2.5535; y_6 = 1.7256; z_6 = -1.0536`
`∴x_7 = 0.2(10 + y_6 – z_6) = 0.2(10 + 1.7256 =1.0536) = 2.5558`
`∴y_7 = 0.57(6 – x6) = 0.5(6 – 2.5535) = 1.7232`
`∴z_7 = -0.2(1 + x_6 + y_6) = -0.2(1 + 2.5535 + 1.7256) = -1.0558`

Iteration 8:
Put `x_7 = 2.5558; y_7 = 1.7232; z_7 = -1.0558`
`∴x_8 = 0.2(10 + y_7 – z_7) = 0.2(10 + 1.7232 =1.0558) = 2.5558`
`∴y_8 = 0.57(6 – x_7) = 0.5(6 – 2.5558) = 1.7221`
`∴_7 = -0.2(1 + x_6 + y_6) = -0.2(1 + 2.5558 + 1.7232) = -1.0558`

Hence, by Gauss Jacobi Iteration Method, the solution is
x = 2.5558, y = 1.7221, z = -1.0558

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Gauss Jacobi Iteration Method
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2017-2018 (December) CBCGS
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