矩陣與數(shù)值分析課程數(shù)值題目

1、1《矩陣與數(shù)值分析》課程數(shù)值實(shí)驗(yàn)題目一、設(shè)，分別編制從小到大和從大到小的順序計(jì)算，????NjNjS2211100S，，并指出有效位數(shù)。10000S1000000S解：程序代碼如下：function[AB]=first(n)fmatlonggT=12(1121n1(n1))k=0S=0fi=2:nS=S1(i^21)endA=Sx=abs(AT)i=0while1ifx1210^(i)i=i1elsebreakendendN1=ki1d

2、isplay(從大到小:)AN1S=0fi=n:1:2S=S1(i^21)endB=Sx=abs(BT)i=0while1ifx1210^(i)i=i1elsebreakendendN2=ki1display(從小到大:)BN23U=triu(A1)Tj=inv(D)(LU)cj=inv(D)btol=1e06k=1fmatlongx=zeros(n1)whilek=Nx(:k1)=Tjx(:k)cjdisp(k)disp(x=)dis

3、p(x(:k1))ifnm(x(:k1)x(:k))toldisp(Theprocedurewassuccessful)disp(Condition||x^(k1)x^(k)||tolwasmetafterkiterations)disp(k)disp(x=)disp(x(:k1))breakendk=k1end結(jié)果輸出TheprocedurewassuccessfulCondition||x^(k1)x^(k)||tolwasmeta

4、fterkiterations60x=0.799998799067310.599998427958700.399998056850090.19999902842505（2）GaussSeidel迭代法程序代碼：clcclearA=[2100121001210012]b=[1000]N=100n=size(A1)D=diag(diag(A))L=tril(A1)U=triu(A1)Tg=inv(DL)Ucg=inv(DL)btol=1e0