第3章_解线性方程组的迭代法_962109547(5)
时间：2025-06-09
hao
阵，f (D L) 1b
x
(k 1)
Gx
(k)
f
例1.3 用Jacobi 迭代法和Gauss-Seidel迭代法解
6 10x1 x2 2x3
x1 11x2 x3 3x4 25
2x x 10x x 11234 1 3x2 x3 8x4 15
此方程组有唯一解 x* (1,2, 1,1)T R4
Jacobi x1
x2x3
(k 1)
1101
x2x1
(k)
15
x3
(k)
x3
35
3x4
(k)
(k 1)(k)
1
(k)
(k 1)
x4
(k 1)
111111111(k)1(k)3(k)11 x1 x2 x4
51010103(k)1(k)15 x2 x3
888
25
k0123
x1
(k)
x2
(k)
x3
(k)
x4
(k)
00.61.04730.9326
02.27271.71592.053
0 1.1000 0.8052 1.0493
01.8750.88521.1309
8910
1.00060.99971.0001
x
(10)
1.99872.00041.9998 x
(10)
(9)
0.9990 1.0004 0.99988.0 10
4
0.99891.00060.9998
3
x
(10)
1.9998
10
事实上，x x
*
0.0002
Gauss-Seidel: x1 x2
(k 1)
(k 1)
110
x2
k(
x3
)
15
x3
k(
)3
5
111
x1
(k 1)
111
(k)
311
x4
(k)
2511
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