Demonstratie - metoda lui Jacobi

Cuprins
Consideram sistemul

Initializare

A = [2,1;5,7];
b = [4;19];
xn = [2;1];

Pregatirea matricelor metodei

M = diag(diag(A))
M = 2×2
2 0 0 7
N = M -A
N = 2×2
0 -1 -5 0

Prima iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn = 2×1
1.500000000000000 1.285714285714286
ea = 0.500000000000000

A doua iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn = 2×1
1.357142857142857 1.642857142857143
ea = 0.357142857142857

Dupa 25 de iteratii

for k=1:23
xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
end
xn
xn = 2×1
1.000002153146738 1.999996924076088
ea
ea = 2.153146738237410e-06
er =ea/norm(xn,inf)
er = 1.076575024850136e-06