實驗名稱：實驗3jacobi迭代法與Gauss-seidel迭代法實驗題目：給定線性方程組Ax=b如下10x一x+2x=6-x+11x一x+3x=25V12342x—x+10x—x=—1112343x—x+8x=15實驗目的：掌握用jacobi迭代法與Gauss-seidel迭代法求解線性方程組的基本步驟。基礎理論：jacobi迭代法基本思路是方程組Ax=b等價于x=Bx+f,然后通過迭代算出方程組的解；Gauss-seidel迭代法主要是通過對A的分解，構造迭代公式，進而迭代算出方程組的解。實驗環境：操作系統：WindowsXP；實驗平臺：matlab實驗過程：程序：方法:jacobi迭代法n=4;A=[10,-1,2,0;-1,11,-1,3;2,-1,10,-1;0,3,-1,8];b=[6,25,-11,15];x0=[0,0,0,0];x0=zeros(n,1);x=x0;epsilon=input('\n精度=');N=input('\n最大迭代次數N=');fprintf('\n%d:',0);fori=1:nfprintf('\%f',x0(i));end%以下是迭代過程fork=1:N%這是第k步迭代，迭代前的向量在x0口中，迭代后的在x口中normal=0;fori=1:nx(i)=b(i);forj=1:nifj=ix(i)=x(i)-A(i,j)*x0(j);endendx(i)=x(i)/A(i,i);temp=abs(x(i)-x0(i));%求范數與迭代在同一個循環中iftemp>normalnormal=temp;%這里用的是無窮范數endend%第i步迭代結束fprintf('\n%d:',k);fori=1:nx0(i)=x(i);%為下一次迭代準備初值fprintf('%f',x(i));%輸出迭代過程endifnormal<epsilonreturn;endendfprintf（'\n\n迭代%d次后仍未求得滿足精度的解\n',N）;結果：精度=口一00001最大迭代次敷昨我0sWarriing:Seehelpspriri-tfforvalidebc^.pesei^_uenceB.^arTiirigsSeeKelp￡=prirrt￡forvslid&scseqij.encesBEarning:Seehelpsprintfforvalidescapeseauences.Warning:Seehelpsprirrtfforvalidescapeseqi±ences.1:0.60000a2.272F27-1.1000001.8750002:1.0472731.715909-0.8052270.88622?3:0.932S3S2.053306-1.0493411.130S814:1.0151991.953696-0.9681090.9738435:0.9889912.011415-1.0102861.0213516:1.0051991.992241-0.9945220.9944347■:0-9981282.002307-1.0019721.0035948:1.0006251.998670-0-9990360-99SS33EhCL9996742.000448-1a0003691aOOOC1910:1.□□□1191.999768-0-9998280.99978611:0.99953422.000085-1.0000681.00010912:1.□□□□221.999S53-0.9999630.9S996D13:0.9999902.000010-1.0000131.00001914:1.0000041.999993-LL9999940.99999215:0.9999982.000003-1.0000021.00000316:1.□□□□□!1.399999-0.S39999D.99Q999_i結果分析：jacobi迭代法解線性方程組迭代的次數較多。方法二：Gauss-seidel迭代法程序:n=4;A=[10,-1,2,0;-1,11,-1,3;2,-1,10,-1;0,3,-1,8];b=[6,25,-11,15];x=[0,0,0,0];epsilon=input('\n精度=');N=input('\n最大迭代次數N=');fprintf('\n%d:',0);fori=1:nfprintf('\%f',x(i));end%以下是迭代過程fork=1:N%這是第k步迭代，迭代前的向量在x0口中，迭代后的在x[]中normal=0;fori=1:nt=x(i);x(i)=b(i);forj=1:nifj=ix(i)=x(i)-A(i,j)*x(j);endendx(i)=x(i)/A(i,i);temp=abs(x(i)-t);%求范數與迭代在同一個循環中iftemp>normalnormal=temp;%這里用的是無窮范數endend%第i步迭代結束fprintf('\n%d:',k);fori=1:nfprintf('%f',x(i));%輸出迭代過程endifnormal<epsilonreturn;endendfprintf('\n\n迭代%d次后仍未求得滿足精度的解\n',N);結果:精度二口.00001最大迭代次數貝二5弓0^:Varning:5已已helpsprintfforvalid已scapes已qu已ncesVTarning:Seehelpsprintffor.Validescapesequences.IWarnirig:Seehelpsprintfforvalidescapesequences.VTarriing:SeehelpsprintfforvalidescapeSequences.1:0.6000002132^27>0.987-2?3-0.8788642；1.0301822.036338-1.0144560.9843413:1.0065852-i'003555-1.0025270.998351&1-0008612.000298-1.0003070.9998505:1.0000