System of Linear Equations (Iterative Methods) Iterative methods ave based on successive Improvement of initial quesses for the sclution ITERCTIVE METHUDI: JACOBI METHOD Step 1. Rewrite the system Step 2. Initialize a value for the unknowns of zeru. tep 3. Perform iterction until the values af the unknown don't diverage anymore. Example: 3xty-z=3 2×t4y tz=フ X-Y+4z= ey Solution: x = 3-y+Z 3 y= 7-2x-Z 4 4-xty Ist Iteration X, = 3-0+O 3 Y.= 7-210)- (0) =175 2,:4-0+O ご 4 2ND ITERAMON: うーy,+ 2,- 3-115 +1 >0.75 3 3 Y2:7-2x,2、:フー2()-1 そz: 4-X,ty、 4-1+1.15 4-1+1.75 =1.1875 %3D
System of Linear Equations (Iterative Methods) Iterative methods ave based on successive Improvement of initial quesses for the sclution ITERCTIVE METHUDI: JACOBI METHOD Step 1. Rewrite the system Step 2. Initialize a value for the unknowns of zeru. tep 3. Perform iterction until the values af the unknown don't diverage anymore. Example: 3xty-z=3 2×t4y tz=フ X-Y+4z= ey Solution: x = 3-y+Z 3 y= 7-2x-Z 4 4-xty Ist Iteration X, = 3-0+O 3 Y.= 7-210)- (0) =175 2,:4-0+O ご 4 2ND ITERAMON: うーy,+ 2,- 3-115 +1 >0.75 3 3 Y2:7-2x,2、:フー2()-1 そz: 4-X,ty、 4-1+1.15 4-1+1.75 =1.1875 %3D
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Answer the following using Jacobi Method. Show complete solutions.
3. 9x1 + 2x2 - 3x3 = 7
x1 + 12x2 - 9x3 = 2
4x1 -6x2 + 14x3 = 1
This is the example that might help..
Transcribed Image Text:Step 3. Perform iterction until the values af
System of Linear Equations (Iterative Methads)
Iterative methads ave based on successive
improvement of initial quesser for the sdlution
ITERATIVE METHUDI : JACORI METHOD
Step 1. Rewrite the system
Step 2. Initialize
a value for the unknouns
of zeru.
tep 3. Perform iterction until the values al
the unknawn dan't divevge anymore.
falut
Example:
A 3xty-Z =3
2×+4y+zニフ
Solution:
そtん-e = メ
Y=7-2x-Z
4
Z:4-xty
スキメーh
: x:0; y=0,2=0
Ist Itercition
X,こ3-0+O
3
Y.= 7-2(0) - (0?
=1.75
Z,?4-04 O
4
2ND ITERAION:
メ2ラーy,ナ21-3-175+l>0.75
3
3
Y2:1-2x,*2フー2-D.1
そz: 4-Xty、 4-1t15
4-1+1.75
=1.1875
Transcribed Image Text:Sheuld oo be less than 0.005,
5 decimdl
3RD ITERATION:
メッ=3- y2t2,
%3D
3-1+1.1875
=1.0625
%3D
Yo:フ-2x, -Z2
7-210.75)-1-1875
こ
1.078125
2,- 4-メetyz 4 - C0-75) +|
%3D
i,0625
4
'Iteration
1
1.75
2.
0-75
|-1875
3
1.0425
1.078125
U -994792 0.953125 1.003906
10625
4
1.016927
1-001628 0-989583
0-995985 0.994141 0.990175
0.09953
7
1-000676
1.00303
As required, the difference of the last iteaticn
1に2、1:んり1こメ
|
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