Let00000a21L美..0aan1n20a12ain00a2nU三...000A=D-L-UD-A=L+U
Let − − − = 0 0 0 0 0 0 1 2 2 1 an an a L − − − = 0 0 0 0 0 0 2 1 2 1 n n a a a U A= D − L −U D− A= L +U
So theiterative process becomes:x(k+1) = D-1(L +U)x(k) + D-1bLet B=D-'(L+U),f =D-'b , thenx(k+1) = B x(k) + f (k=0,1,2,..)This method is called the Jacobiiteration (J method)for solving the linear equations system.B is the iteration matrix of the Jacobi iterations.Equivalent linear equations system:x=Bx+f←Ax= b
Equivalent linear equations system: x B x f = + Ax = b This method is called the Jacobi iteration (J method) for solving the linear equations system. So the iterative process becomes: x D L U x D b (k 1) 1 (k ) 1 ( ) + − − = + + ( 1) ( ) k k x B x f + = + (k = 0,1,2, ) 1 1 B D L U f D b ( ), − − Let = + = , then B is the iteration matrix of the Jacobi iterations
In matrix form :Ax=b (D-L-U)x=b-UA= Dx=(L+U)x+bD-L台 x=D-(L+U)x+D-"b1fBx(k+1) = B x(k) + fJacobiiterationmatrix
In matrix form： A = -L -U D ( ) ( ) Ax b D L U x b Dx L U x b = − − = = + + 1 1 x D L U x D b ( ) = + + − − B f Jacobi iteration matrix ( 1) ( ) k k x B x f + = +
Solve theeguations system by the Jacobiiteration.Example1and guarantee the error no greater than le-4.820-32X143311-1X221214X3Solutions:82-3411-1A=21400003280000010-4U=L=0011D=0000-2一1004
Example 1 Solve the equations system by the Jacobi iteration, and guarantee the error no greater than 1e-4. = − − 12 33 20 2 1 4 4 11 1 8 3 2 3 2 1 x x x Solutions: − − = 2 1 4 4 11 1 8 3 2 A = 0 0 4 0 11 0 8 0 0 D − = 0 0 0 0 0 1 0 3 2 U − − = − 2 1 0 4 0 0 0 0 0 L
3-8141040B =D-I(L+U)-11-2104(2.5)f = D-1b=33Take initial value x(0) =[0 0 O}’', and use the Jacobiiteration:x(k+1) = Bx(k) + f (k =0,1,2,..n,...)
1 B D L U ( ) − = + − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 f D b −1 = = 3 3 2.5 ( 1) ( ) k k x Bx f + = + (k = 0,1,2, n, ) (0) [0 0 0]T Take initial value x = , and use the Jacobi iteration: