TU Wien:Computernumerik VU (Schranz-Kirlinger)/Übungen SS16/Beispiel 15

Lösung[Bearbeiten | Quelltext bearbeiten]

Matlab Code

function Bsp5

%% a)

Aex =[1.253672417 1.247798111 ; -2.672344812 2.695328007];

bex =[3.654199872 2.479981003]';

xex = Aex\bex;

fprintf('%s %0.15f %s %0.15f\n','correct x1 = ', xex(1), 'x2 = ', xex(2));

Af=[1.253672000 1.247798000 ; -2.672344000 2.695328000];

bf=[3.654199000 2.479981000]';

xf=Af\bf;

fprintf('%s %0.15f %s %0.15f\n','false x1 = ', xf(1), 'x2 = ', xf(2));

%kappa_2 and kappa_inf

kappa2 = cond(Aex,2);

kappainf = cond(Aex,inf);

fprintf('%s %0.15f %s %0.15f\n', 'kappa_2 = ', kappa2, 'kappa_inf = ', kappainf);

%% b) Aex=[1.743681226 -0.5287326143 ; 4.359203065 -1.321302803];

bex=[2.666771987 6.667195145]';

xex=Aex\bex;