) 中图学技术大荸学系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS A=1 1、 Jacobi迭代 01/2-1/2 B=D(L+U)=-10 1/21/20 特征值为-B=23+02=0 1=0,223=±i 2、 Gauss-Sede迭代 01/2-1/2 B=(D-L)U=0-1/2-1/2 00 1/2 x1=0,3=-2数 学 系 University of Science and Technology of China DEPARTMENT OF MATHEMATICS 1 0 1/ 2 1/ 2 ( ) 1 0 1 1/ 2 1/ 2 0 B D L U −   −   = + = − −       1、Jacobi迭代 2 1 1 1 1 1 1 1 2 A   −   =       − 特征值为 3 5 0 4    I B− = + = 1 2,3 5 0, 2   = =  i 1 0 1/ 2 1/ 2 ( ) 0 1/ 2 1/ 2 0 0 1/ 2 B D L U−   −   = − = − −       − 2、Gauss－Siedel迭代 1 2,3 1 0, 2   = = −