Gauss Seidel

David the Mathlete · **Joined:** Fri, 8 Apr 2005 16:12:09 UTC **Posts:** 333

Hi, I'm trying to do the matrix method for Gauss Seidel and the iterations are coming out different than the equation method so i wanted to know if anyone could explain to me how to go about doing the gauss seidel method with matrices i have

-2x_1+x_2=-200

so I created the strictly upper matrix and the lower matrix and got

Lower:
$-2\ \ 0\ \ 0$
$1\ \ -2\ \ 0$
$0\ \ 1\ \ -2$

Strictly Upper
$0\ \ 1\ \ \ 0$
$0\ \ 0\ \ 1$
$0\ \ 0\ \ 0$

then I found the inverse of the Lower matrix
$-1/2\ \ \ \ \ 0\ \ \ \ \ \ \ \ 0$
$-1/4\ \ -1/2\ \ \ \ 0$
$-1/8\ \ -1/4\ \ -1/2$

then i found Linverse*U and Linverse*b

then used the iterative method of

$x^{k+1}=-L^{-1}Ux^k+L^{-1}b$

and for my first iteration i got

x_1=-25

with an initial guess of

x_1=x_2=x_3=50

and the solution vector is
x_1=125

outermeasure · **Posted:** Thu, 30 Sep 2010 07:06:24 UTC

David the Mathlete wrote:

Hi, I'm trying to do the matrix method for Gauss Seidel and the iterations are coming out different than the equation method so i wanted to know if anyone could explain to me how to go about doing the gauss seidel method with matrices i have

-2x_1+x_2=-200

so I created the strictly upper matrix and the lower matrix and got

Lower:
$-2\ \ 0\ \ 0$
$1\ \ -2\ \ 0$
$0\ \ 1\ \ -2$

Strictly Upper
$0\ \ 1\ \ \ 0$
$0\ \ 0\ \ 1$
$0\ \ 0\ \ 0$

then I found the inverse of the Lower matrix
$-1/2\ \ \ \ \ 0\ \ \ \ \ \ \ \ 0$
$-1/4\ \ -1/2\ \ \ \ 0$
$-1/8\ \ -1/4\ \ -1/2$

then i found Linverse*U and Linverse*b

then used the iterative method of

$x^{k+1}=-L^{-1}Ux^k+L^{-1}b$

and for my first iteration i got

x_1=-25

with an initial guess of

x_1=x_2=x_3=50

and the solution vector is
x_1=125

Gauss-Seidel is an iterative method, and there are no reason to expect you arrive at the exact solution with just one iteraton.

In your example, the matrices are
$\begin{aligned} A&=\begin{pmatrix}-2&1&0\\1&-2&1\\0&1&-2\end{pmatrix}\\ L_*&=\begin{pmatrix}-2&0&0\\1&-2&0\\0&1&-2\end{pmatrix}\\ U_*&=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\\ \mathbf{b}&=\begin{pmatrix}-200\\0\\-100\end{pmatrix} \end{aligned}$
and the iterative scheme is
$\mathbf{x}^{(k+1)}=L_*^{-1}(\mathbf{b}-U_*\mathbf{x}^{(k)})$
i.e.
$\begin{aligned} x^{(k+1)}_1&=-\frac{1}{2}\left(-200-x^{(k)}_2\right)&&=\frac{1}{2}\left(200+x^{(k)}_2\right)\\ x^{(k+1)}_2&=-\frac{1}{2}\left(-x^{(k)}_3-x^{(k+1)}_1\right)&&=\frac{1}{2}\left(x^{(k)}_3+x^{(k+1)}_1\right)\\ x^{(k+1)}_3&=-\frac{1}{2}\left(-100-x^{(k+1)}_2\right)&&=\frac{1}{2}\left(100+x^{(k+1)}_2\right) \end{aligned}$
So the iterative scheme yields
$\begin{tabular}{c||c|c|c} $k$&$x^{(k)}_1$&$x^{(k)}_2$&$x^{(k)}_3$\\\hline 0&50&50&50\\ 1&125&87.5&93.75\\ 2&143.75&118.75&109.375\\ 3&159.375&134.375&117.1875\\ 4&167.1875&142.1875&121.09375\\ 5&171.09375&146.09375&123.046875\\ 6&173.046875&148.046875&124.0234375\\ \vdots&\vdots&\vdots&\vdots \end{tabular}$

Compare this with the exact solution
$\mathbf{x}=\begin{pmatrix}175\\150\\125\end{pmatrix}$

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Gauss Seidel

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