\[A \mathbf{x}= \mathbf{b}\]
numerically. Both methods use iteration formulae derived from the equations. The difference lies in the initial solution. The Jacobi method sets all \[x_i =0\]
and the Gauss - Seidel method sets \[x_i = b_i/a_{ii}\]
Suppose we have the system of equations
\[14x_1+2x_2+4x_3=-10\]
\[16x_1+40x_2-4x_3=55\]
\[-2x_1+4x_2-16x_3=-38\]
Rearrange each equation for
\[x_1, \: x_2, \: x_3\]
respectively.\[x_1=-10/14-2/14x_2-4/14x_3\]
\[x_2=55/40-16/40x_1+4/40x_3\]
\[x_3=-38/-16-2/16x_1+4/16x_2\]
Now use the iteration formulae
\[x^{(n+1)}_1=-10/14-2/14x^{(n)}_2-4/14x^{(n)}_3=-0.714-0.143x^{(n)}_2-0.286x^{(n)}_3\]
\[x_2=55/40-16/40x^{(n)}_1+4/40x^{(n)}_3=1.375-0.400x^{(n)}_1+0.100x^{(n)}_3\]
\[x_3=-38/-16-2/16x^{(n)}_1+4/16x^{(n)}_2=2.275-0.125x^{(n)}+0.250x^{(n)}_2\]
Take initial solution
\[x^{(0)}_1=-10/14=-0.714, \: x^{(0)}_2=55/40=1.375, \: x^{(0)}_3=38/16=2.375\]
Then
\[x^{(1)}_1=-0.714-0.143 \times 1.375-0.286 \times 2.375=-1.590\]
\[x^{(1)}_2=1.375-0.400 \times -0.714+0.100 \times 2.375=1.899\]
\[x^{(1)}_3=2.375-0.125 \times -0.714 +0.250 \times 1.375=2.808\]
Continuing in this way gives the table.
| nbsp; \[n\] |
nbsp; \[x^{(n)}_1\] |
nbsp; \[x^{(n)}_2\] |
nbsp; \[x^{(n)}_3\] |
| 0 | -0.714 | 1.375 | 2.375 |
| 1 | -1.599 | 1.899 | 2,808 |
| 2 | -1.789 | 2.293 | 3.049 |
| 3 | -1.914 | 2.396 | 3.172 |
| 4 | -1.964 | 2.458 | 3.213 |
| 5 | -1.984 | 2.482 | 3.236 |
| 6 | -1.994 | 2.493 | 3.244 |
| 7 | -1.998 | 2.498 | 3.247 |
\[x_1=-2, \: x_2=2.5, \: x_3=3.25\]