Demonstratie - metoda Gauss-Seidel

Contents

Consideram sistemul

$$\left[ \begin{array} [c]{cc}
2 & 1\\
5 & 7
\end{array} \right]x = \left[
\begin{array}
[c]{c}
4\\
19
\end{array}
\right]
  $$

Initializare

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

Pregatirea matricelor metodei

$$M = tril(A), N = M - A$$

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

Prima iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn =
   1.500000000000000
   1.642857142857143
ea =
   0.642857142857143

A doua iteratie

xv = xn;
xn = M\(N*xv+b);
ea = norm(xn-xv,inf);
xn, ea
xn =
   1.178571428571429
   1.872448979591837
ea =
   0.321428571428571

Dupa 20 de iteratii

for k=1:19
    xv = xn;
    xn = M\(N*xv+b);
    ea = norm(xn-xv,inf);
end
xn
ea
er =ea/norm(xn,inf)
xn =
   1.000000000569914
   1.999999999592918
ea =
     1.025845408619830e-09
er =
     5.129227044143157e-10