数值分析实习报告- 1 - 数值分析实习报告- 2 - 上机实习题一一、题目：已知 A 与 b 12.38412,2.115237,-1.061074,1.112336,-0.113584,0.718719,1.742382,3.067813,-2.031743 2.115237,19.141823,-3.125432,-1.012345,2.189736,1.563849,-0.784165,1.112348,3.123124 -1.061074,-3.125432,15.567914,3.123848,2.031454,1.836742,-1.056781,0.336993,-1.010103 1.112336,-1.012345,3.123848,27.108437,4.101011,-3.741856,2.101023,-0.71828,-0.037585 A= -0.113584,2.189736,2.031454,4.101011,19.897918,0.431637,-3.111223,2.121314,1.784137 0.718719,1.563849,1.836742,-3.741856,0.431637,9.789365,-0.103458,-1.103456,0.238417 1.742382,-0.784165,-1.056781,2.101023,-3.111223,-0.103458,14.7138465,3.123789,-2.213474 3.067813,1.112348,0.336993,-0.71828,2.121314,-1.103456,3.123789,30.719334,4.446782 -2.031743,3.123124,-1.010103,-0.037585,1.784317,0.238417,-2.213474,4.446782,40.00001 b={2.1874369,33.992318,-25.173417,0.84671695,1.784317,-86.612343,1.1101230,4.719345,-5.6784392} 1．用 Household 变换，把 A 化为三对角阵B（并打印 B）。2．用超松弛法求解BX=b （取松弛因子 ω =1.4,X(0)=0,迭代 9 次）。3．用列主元素消去法求解BX=b 。二、解题方法的理论依据: 1 、用 Householder 变换的理论依据﹝1﹞ 令 A0=A,a(ij)1=a(ij),已知 Ar_1 即 Ar_1=a(ij)r ﹝2﹞ Sr=sqrt(pow(a,2)) ﹝3﹞ a(r)=Sr*Sr+abs(a(r+1,r))*Sr ﹝4﹞ y(r)=A(r_1)*u?/a?﹝5﹞ Kr=(/2)*Ur的转置 *Yr/a?﹝6﹞ Qr=Yr-Kr*Ur ﹝7﹞ Ar=A(r-1)-(Qr*Ur的转置 +Ur*Qr 的转置 ) r=1,2,,⋯⋯,n -2 2 、用超松弛法求解其基本思想：在高斯方法已求出x(m),x(m-1)的基础上，组合新的序列，从而加快收敛速度。其算式：][][0][0][~][]][[][]1[0]][[]1][[]1[]][[]1][[][~iXiXiXiXwiXiiBibiXiiBiiBiXiiBiiBiX其中 ω 是超松弛因子，当ω >1 时，可以加快收敛速度3 、用消去法求解用追赶消去法求Bx=b 的方法：][]1[1ibid , ]][1[]2[1iiBia , ][]1[]1][[][]][[]1[]1][[ibiXiiBiXiiBiXiiB数值分析实习报告- 3 - ]][[]1[1iiBib , ]1][[]1[1iiBic , q1[0]=0 , u1[0]=0 , 8,,2,1]),[1][1][1(][1][1iiqiaibiciq9,,2,1]),1[1][1][1(])[1][1][1(}[1iiqiaibiuiaidiu x[9]=u1[9] 1,,7,8...
