Solve the given system of linear equations using Gauss-Seidel Method. The answer should have atleast 5 decimal places.Box thw final answer. 1.) 3x1 - x2 + x3 = 2 2x1 - 3x2 - x3 = 7 3x1 - x2 + 4x3 = 5 This is the example that might help.
Solve the given system of linear equations using Gauss-Seidel Method. The answer should have atleast 5 decimal places.Box thw final answer. 1.) 3x1 - x2 + x3 = 2 2x1 - 3x2 - x3 = 7 3x1 - x2 + 4x3 = 5 This is the example that might help.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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Question
Solve the given system of linear equations using Gauss-Seidel Method. The answer should have atleast 5 decimal places.Box thw final answer.
1.) 3x1 - x2 + x3 = 2
2x1 - 3x2 - x3 = 7
3x1 - x2 + 4x3 = 5
This is the example that might help.
Transcribed Image Text:3rd iteration:
X₁ =
1/₂ =
X₂²
4th:
X₁
X₂
XL L LL FOX
3-1015625 +1.019531.
3
7-2(1.001302)-1.019331
4
= 0.994466
9-1001 302 +0.994466 = 0.998291₂
4
3-1.001302 +0.998291
3
=1061275
7-2(1.001275) -0.998291
= 0.999790
X3
3-10156662811 0-4.00 136 2
X₁ =
41
7-2(1.001302)-1.019531
$
= 0.994466~
9-1001 302 +0.994466 = 0.998291₂
4
3-1.001302 +0.998291
3
=1.061275
7-2(1001275) -0.998291
4
=0.999790
4.1.001 275 +0.999790
-=0.997679
4
xz
x3
1/₂ =
X₂ =
4th:
X₁ =
X₂
X3
X₁
=1.001302
ITER
O
O
D
O
1
1.25
10625
2
3
0.9375 1.015625 1.019531
1.001302 0.994 466 0.998291.
4 1.001275 0.999790 10.999629-
X₁² |
X₂ = 1; X3 =1
.
Transcribed Image Text:ITERATIVE METH. L&L 9 XEIDEL METHOD
Step 1: Rewrite the system
Step2: Get the first approximation.
Step 3: Perform iteration until the desired accuracy is obtained.
NOTE: The difference of the iteration should be less than 0.005.
Ex 1) 3x₁ + x₂ −X₂ = 3
ITER
X₁
O
2x + 4x₂ + x₂ =7
1
X₁
X₂ +4×3=4
2
3
4
3-X₂ + X₂
3
SOL' Ni
X₁ =
2
X₁ =
• 3 - xy + f₂ ²
3
X₂ = 7-2x₁-xx
4
x₂-4-x₁ + x²
X3
Ist iteration:
X₁ = 3-D + D = 1₁
X₁₂ = 3-2 (1) +0-125,
4
4-1+125
X₂ = 7 - 2 (1) 46 - 25² 11 9×
4
X3=
4-1+125
4
= 1.0625
2nd Iteration:
X₁ =
3-1.25 +1.0625
3
X₂= 7-2 (0.9375)-1.0625
4
1.015625
X3 = 4-0.9375 +1.015625 -=1.0195311
3rd iteration:
X₂
JOLEIOS LLAKEN
SOL' Ni
LLLEBQX
= 0.9375
Xz
D
1
1.25
0.9375 1.015625
1.001302 0.99446
1001275 0.999796
X₁ = |
1.001302 0.99
1.001275 0.49
X₁=1 X₂
4
O
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