3RD ITERATION: x₂ = 3-X₂ +2₂ 3-1 + 1.1875 = 1.0625 3 3 Yo= 7-2x₂₁-22-7-2 10.757-1-1875 4 4 1.078125 F 23 = 4-X₂+x₂ - 4 - (0-75) +1 H = 1.0625 4 9 Iteration Y O 1 2 O Z 5 decimal O O 1 1.75 1 0.75 1 1-1875 3 1.0625 1-078125 1.0625 4 0.994792 0.953125) 1-003906 S 1.016927 1.001628 0-989583 6 0-995985 0.994141 0.996175 0.999539 1-000678 1.00303 As required, the difference of the last iteration Should be less than 0.005, X=1; y = 1 ; 2=

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Solve the given system of linear equations using the Jacobi 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/ guide that might help.

 

L
3RD ITERATION:
x₂ = 3-√₂+² 2 = 3-1 + 1.1875
3
3
Yo = 7-2x₂-2₂
- 2x2-2₂
7-210-757-1-1875
4
4
1.078125
Z₂
2
4-x₂+x2= 4 -(0-75) +1 1.0625
4
។
Iteration
Y
Z
O
1
1.75
1
0.75
1
1-1875
1.0625
1.078125
1.0625
0.9947921 0.953125 1.003906
5
6
1.016927 1-001628 0-989583
0-995985 | 0.994141 0.996175
1.00303
0.999539
1-000678
As required, the difference of the last iteration
Shoulder be less than 0.005,
| X=1; y = 1 ; 2=1
2
3
4
(1
O
5 decimal
O
= 1.0625
Transcribed Image Text:L 3RD ITERATION: x₂ = 3-√₂+² 2 = 3-1 + 1.1875 3 3 Yo = 7-2x₂-2₂ - 2x2-2₂ 7-210-757-1-1875 4 4 1.078125 Z₂ 2 4-x₂+x2= 4 -(0-75) +1 1.0625 4 ។ Iteration Y Z O 1 1.75 1 0.75 1 1-1875 1.0625 1.078125 1.0625 0.9947921 0.953125 1.003906 5 6 1.016927 1-001628 0-989583 0-995985 | 0.994141 0.996175 1.00303 0.999539 1-000678 As required, the difference of the last iteration Shoulder be less than 0.005, | X=1; y = 1 ; 2=1 2 3 4 (1 O 5 decimal O = 1.0625
System of Linear Equations (Iterative Methods)
Iterative methods are based on successive
improvement of initial guesses for the solution.
ITERATIVE METHODI: JACOBI METHOD
Step 1. Rewrite the system
Step 2. Initialize a value for the unknowns
of zero.
Step 3. Perform iteration until the values of
the unknown don't diverge anymore.
folut
Example &
肉丸 3x+2y2=3
4y+z=7
X-y + 4z = 4
Solution:
NAA
x = 3-y+z
3
Y
= 7-2x-z
4
Z=
4-x+x
4
1st Iteration : x=0; y=0,2 = 0
X, = 3-ото
= 1
Y₁ = 7-2(0) - (0)
= 1.75
2,4-04 으
= 1
4
2ND ITERATION:
x₂ = 3-y₁ +2₁= 3 ~ 1.75 +1 = 0.75
3
3
Y2:7-2x, 7,7-20)-(1
4
4
Z₂ = 4-X₁tY₁ = 4-1+1.75
4
4
= 1.1875
Transcribed Image Text:System of Linear Equations (Iterative Methods) Iterative methods are based on successive improvement of initial guesses for the solution. ITERATIVE METHODI: JACOBI METHOD Step 1. Rewrite the system Step 2. Initialize a value for the unknowns of zero. Step 3. Perform iteration until the values of the unknown don't diverge anymore. folut Example & 肉丸 3x+2y2=3 4y+z=7 X-y + 4z = 4 Solution: NAA x = 3-y+z 3 Y = 7-2x-z 4 Z= 4-x+x 4 1st Iteration : x=0; y=0,2 = 0 X, = 3-ото = 1 Y₁ = 7-2(0) - (0) = 1.75 2,4-04 으 = 1 4 2ND ITERATION: x₂ = 3-y₁ +2₁= 3 ~ 1.75 +1 = 0.75 3 3 Y2:7-2x, 7,7-20)-(1 4 4 Z₂ = 4-X₁tY₁ = 4-1+1.75 4 4 = 1.1875
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