Question

Solve Problem 6.2 using the Jacobi iterative method. Start with ${\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{x}}_{\mathbf{2}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ and continue until (6.2.2) is satisfied with $\mathit{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.01}$.

Short answer

The solution of the system of equation using Jacobi method is$x_1=0.4243,x_2=0.2325and{x}_3=-0.0152$

Step-by-step solution

Given data

$\begin{array}{l}{x}_1\left(1\right)=0\\ x_2\left(1\right)=0\\ x_3\left(1\right)=0\\ x_4\left(1\right)=0\end{array}$

And

$\epsilon =0.01$

Determine matrix

Consider the following equation.

$\begin{array}{c}6x_1+2x_2+x_3=3\mathrm{......}\left(1\right)\\ 4x_1+10x_2+2x_3=4\mathrm{......}\left(2\right)\\ 3x_1+4x_2+14x_3=2\mathrm{......}\left(3\right)\end{array}$

Write the above equations in matrix form.

$\left[\begin{array}{ccc}\begin{array}{l}6\\ 4\\ 3\end{array}& \begin{array}{l}2\\ 10\\ 4\end{array}& \begin{array}{l}1\\ 2\\ 14\end{array}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right]$

Use Jacobi method

Write the expression using Jacobi method.

$x\left(i+1\right)=Mx\left(i\right)+D^{-1}B\mathrm{......}\left(4\right)$

From equation (4), A becomes

$A=\left[\begin{array}{ccc}6& 2& 1\\ 4& 10& 2\\ 3& 4& 14\end{array}\right]$

Hence, the matrix consisting of the diagonal elements of A becomes:

$D=\left[\begin{array}{ccc}6& 0& 0\\ 0& 10& 0\\ 0& 0& 14\end{array}\right]$

The inverse of D becomes

$\begin{array}{c}{D}^{-1}={\left[\begin{array}{ccc}6& 0& 0\\ 0& 10& 0\\ 0& 0& 14\end{array}\right]}^{-1}\\ =\left[\begin{array}{ccc}\frac{1}{6}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{14}\end{array}\right]\end{array}$

Determine matrix M

$\begin{array}{c}M=D^{-1}\left(D-A\right)\\ =\left[\begin{array}{ccc}\frac{1}{6}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{14}\end{array}\right]\left(\left[\begin{array}{ccc}6& 0& 0\\ 0& 10& 0\\ 0& 0& 14\end{array}\right]-\left[\begin{array}{ccc}6& 2& 1\\ 4& 10& 2\\ 3& 4& 14\end{array}\right]\right)\\ =\left[\begin{array}{ccc}0& -0.3333& -0.1667\\ -0.4& 0& -0.2\\ -0.2143& -0.2857& 0\end{array}\right]\end{array}$

Substitute the value of D and M in equation (4).

$\left[\begin{array}{c}{x}_1\left(i+1\right)\\ x_2\left(i+1\right)\\ x_3\left(i+1\right)\end{array}\right]=\left[\begin{array}{ccc}0& -0.3333& -0.1667\\ -0.4& 0& -0.2\\ -0.2143& -0.2857& 0\end{array}\right]\left[\begin{array}{c}{x}_1\left(i\right)\\ x_2\left(i\right)\\ x_3\left(i\right)\end{array}\right]+\left[\begin{array}{ccc}\frac{1}{6}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{14}\end{array}\right]\mathrm{......}\left(5\right)$

Use initial condition

Use the initial conditions$x_1\left(0\right)=0,x_2\left(0\right)=0$ and$x_3\left(0\right)=0$ in the equation (5).

$\begin{array}{c}{x}_1\left(1\right)=-0.3333x_2\left(0\right)-0.1667x_3\left(0\right)+0.1667\times 3\\ =0.5001\\ x_2\left(1\right)=-0.4x_1\left(0\right)-0.2x_3\left(0\right)+\left(0.1\times 4\right)\\ =0.4\\ x_3\left(1\right)=-0.2143x_1\left(0\right)-0.2857x_2\left(0\right)+0.0714\times 2\\ =0.1428\end{array}$

Determine iteration table

Repeat the above steps to get the value of $x_1\left(n\right),x_2\left(n\right)$and $x_3\left(n\right)$till $\epsilon <0.01$.

Draw the iteration table.

n	0	1	2	3	……	9	10
$x_1\left(n\right)$	0	0.5001	0.3430	0.4561	..	0.4258	0.4243
$x_2\left(n\right)$	0	0.4	0.1714	0.2785	..	0.2340	0.2325
$x_3\left(n\right)$	0	0.1428	-0.0787	-0.0203	..	-0.0140	-0.0152
$\epsilon$	inf	0.5715	0.6248	0.2334	..	0.006

After the iteration the value of is the less than 0.01, thus stop the iteration.

Therefore, the solution of the system of equation using Jacobi method is

$x_1=0.4243,x_2=0.2325and{x}_3=-0.0152$