Assume we have the following a system of linear equation I1 – 8x2 – 2x3 = 1 I1 +22 + 5x3 = 4 3x1 – 12 + x3 =-2 If we rearrange the equation to form a strictly diagonally dominant system, we have 3x1 – 12 + x3 =-2 I1 – 8r2 – 2x3 = 1 Ti + 12 + 5x3 = 4 | | Note that a strictly diagonally dominant system guarantee the convergence of the methods such as Jacobi or Gauss-Seidal Method. If we apply two steps of the Gauss-Seidal Methods from an initial guess „(1) „(1) „(1). (3) (3) (3), (x" , r," , x") = (0,0,0), then (x", x", (3) (3) O (2", 12, 3 ) = (- 180 191 361 89 720' 80 (3) (3) (3) 島,) 49 40 120 24
Assume we have the following a system of linear equation I1 – 8x2 – 2x3 = 1 I1 +22 + 5x3 = 4 3x1 – 12 + x3 =-2 If we rearrange the equation to form a strictly diagonally dominant system, we have 3x1 – 12 + x3 =-2 I1 – 8r2 – 2x3 = 1 Ti + 12 + 5x3 = 4 | | Note that a strictly diagonally dominant system guarantee the convergence of the methods such as Jacobi or Gauss-Seidal Method. If we apply two steps of the Gauss-Seidal Methods from an initial guess „(1) „(1) „(1). (3) (3) (3), (x" , r," , x") = (0,0,0), then (x", x", (3) (3) O (2", 12, 3 ) = (- 180 191 361 89 720' 80 (3) (3) (3) 島,) 49 40 120 24
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Assume we have the following a system of linear equation
x1 – 8x2 – 2x3 = 1
X1 + x2 + 5x3 = 4
3x1 – x2 + x3 =-2
%3D
If we rearrange the equation to form a strictly diagonally dominant system, we have
3x1 – x2 + x3 = -2
X1 – 8x2 – 2x3 = 1
21 + x2 + 5x3 = 4
|
Note that a strictly diagonally dominant system guarantee the convergence of the methods such as
Jacobi or Gauss-Seidal Method.
If we apply two steps of the Gauss-Seidal Methods from an initial guess
(1) „(1)
(x", x" , a") :
(1),
= (0,0,0), then (", x"
(3) (3) „(3)
", x"):
= ?
(3) (3)
361 89
720' 80
191
O (", x", r) = (-
180
O (", 2,", a") = (-;
(3) (3) „(3)-
39
49
23
120' 24
40
O (2), =°, =9) = (-- )
(3) (3)
, x", a) = (- 39
47
120' 22
O (2", x", ") = (- 190
171
(3) (3) „(3)-
359 87
720' 80
(3) (3)
(3)
O (2", a", x") = (0,0, 0)
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