## Solving Systems of Simultaneous Linear Equations By Jacobi Iteration

Solving large systems of simultaneous linear equations, especially with non integer coefficients is long and arduous, with lots of opportunities for error. There is a method, especially suitable for use with computers which involves iteration. It is called Jacobi iteration.
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
$x^{(n+1)}_1=-10/14-2/14x^{(n)}_2-4/14x^{(n)}_3=-0.714-0.143x^{(n)}_2-0.286x^{(n)}_3$

$x^{(n+1)}_2=55/40-16/40x^{(n)}_1+4/40x^{(n)}_3=1.375-0.400x^{(n)}_1+0.100x^{(n)}_3$

$x^{(n+1)}_3=-38/-16-2/16x^{(n)}_1+4/16x^{(n)}_2=2.275-0.125x^{(n)}+0.250x^{(n)}_2$

Take initial solution
$x^{(0)}_1=x^{(0)}_2=x^{(0)}_3=0$

Then
$x^{(1)}_1=-0.714, \: x^{(1)}_2=1.375, \: x^{(1)}_3=2.375$

$x^{(2)}_1=-0.714-0.143 \times 1.375-0.286 \times 2.375=-1.590$

$x^{(2)}_2=1.375-0.400 \times -0.714+0.100 \times 2.375=1.899$

$x^{(2)}_3=2.375-0.125 \times -0.714 +0.250 \times 1.375=2.808$

Continuing in this way gives the table.
 $n$ $x^{(n)}_1$ $x^{(n)}_2$ $x^{(n)}_3$ 0 0 0 0 1 -0.714 1.375 2.375 2 -1.599 1.899 2,808 3 -1.789 2.293 3.049 4 -1.914 2.396 3.172 5 -1.964 2.458 3.213 6 -1.984 2.482 3.236 7 -1.994 2.493 3.244 8 -1.998 2.498 3.247
The iterations are converging. In fact the true solution is
$x_1=-2, \: x_2=2.5, \: x_3=3.25$