Question

Repeat Problem 6.6 using the Gauss-Seidel iterative method.Which

method converges more rapidly?

Short answer

The solution of the system is $x_1=0.222803,x_2=0.533775and{x}_3=-0.58138$.

The Gauss-Seidel method converges more rapidly than Jacobean method.

Step-by-step solution

Write the given data from question.

The initial conditions are

$$\begin{gathered} x_1\left(0\right)=x_2\left(0\right) \\ {x}_1\left(0\right)=x_3\left(0\right) \\ {x}_1\left(0\right)=0 \end{gathered}$$

The matrix for the system,

$\left[\begin{array}{ccc}8& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right]$

The specific tolerance level, $\epsilon =0.01$.

Determine the equation to find the solution of the system.

The general equation of the system is given as follows.

${\mathbf{A}}_{\mathbf{X}}\mathbf{=}\mathbf{y}$ …… (1)

Here x,yare the N vectors and Ais the $\mathbf{N}\mathbf{\times }\mathbf{N}$square matrix.

The equation to calculate the matrix M of Jacobi iteration method is given as

follows.

$\mathbf{M}\mathbf{=}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\left(D-A\right)$ …… (2)

Here Dis the matrix consisting of the diagonal elements of Amatrix.

$\mathbf{x}\left(i+1\right)\mathbf{=}\mathbf{Mx}\left(i\right)\mathbf{+}{\mathbf{D}}^{\mathbf{-}\mathbf{1}}\mathbf{y}$ …… (3)

The equation for the Jacobi method as,

Calculate the solution for the system.

Compare the given system equation with equation (1).

$\mathrm{A}=\left[\begin{array}{ccc}8& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right],\mathrm{x}=\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right],\mathrm{y}=\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right]$

Convert the matrix A into a lower triangular matrix as D .

$\mathrm{D}=\left[\begin{array}{ccc}8& 0& 0\\ 4& 6& 0\\ 3& 4& 14\end{array}\right]$

Calculate the inverse of the D matrix.

${\mathrm{D}}^{-1}=\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]$

Calculate the matrix M for Jacobi method,$$\begin{gathered} \mathrm{Substitude}\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]\mathrm{for}{\mathrm{D}}^{-1},\left[\begin{array}{ccc}8& 0& 0\\ 4& 6& 0\\ 3& 4& 14\end{array}\right]\mathrm{for}\mathrm{D}\mathrm{and} \\ \left[\begin{array}{ccc}8& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right]\mathrm{for}\mathrm{A}\mathrm{into}\mathrm{equation}\left(2\right). \end{gathered}$$

$$\begin{gathered} \mathrm{M}=\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]\left(\left[\begin{array}{ccc}8& 0& 0\\ 4& 6& 0\\ 6& 4& 14\end{array}\right]-\left[\begin{array}{ccc}8& 2& 1\\ 4& 6& 2\\ 3& 4& 14\end{array}\right]\right) \\ \mathrm{M}=\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]\left[\begin{array}{ccc}0& -2& -1\\ 0& 0& -2\\ 0& 0& 0\end{array}\right] \\ \mathrm{M}=\left[\begin{array}{ccc}0& 0.125& -0.125\\ 0& 0.166& -0.2487\\ 0& -0.00595& 0.09817\end{array}\right] \end{gathered}$$

Expressed the equation of the system,

$\mathrm{Substitude}\left[\begin{array}{ccc}0& -0.25& -0.125\\ 0& 0.1666& -0.2487\\ 0& -0.00595& 0.09817\end{array}\right]\mathrm{for}M,\left[\begin{array}{cc}{x}_1& \left(i\right)\\ x_2& \left(i\right)\\ x_3& \left(i\right)\end{array}\right]\mathrm{for}x\left(i\right),$

$\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]\mathrm{for}{\mathrm{D}}^{-1},\mathrm{and}\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right]\mathrm{for}\mathrm{y}\mathrm{into}\mathrm{equation}\left(3\right).$

$$\begin{gathered} \left[\begin{array}{cc}{x}_1& \left(i+1\right)\\ x_2& \left(i+1\right)\\ x_3& \left(i+1\right)\end{array}\right]=\left[\begin{array}{ccc}0& -0.25& -0.125\\ 0& 0.1666& -0.2487\\ 0& -0.00595& 0.09817\end{array}\right]\left[\begin{array}{cc}{x}_1& \left(i\right)\\ x_2& \left(i\right)\\ x_3& \left(i\right)\end{array}\right]+\left[\begin{array}{ccc}0.125& 0& 0\\ -0.0833& 0.166& 0\\ -0.0029& -0.0476& 0.0714\end{array}\right]\left[\begin{array}{c}3\\ 4\\ 2\end{array}\right] \\ \left[\begin{array}{cc}{x}_1& \left(i+1\right)\\ x_2& \left(i+1\right)\\ x_3& \left(i+1\right)\end{array}\right]=\left[\begin{array}{ccc}0& -0.25& -0.125\\ 0& 0.1666& -0.2487\\ 0& -0.00595& 0.09817\end{array}\right]\left[\begin{array}{cc}{x}_1& \left(i\right)\\ x_2& \left(i\right)\\ x_3& \left(i\right)\end{array}\right]+\left[\begin{array}{c}0.375\\ 0.4141\\ -0.5652\end{array}\right] \end{gathered}$$

Separate the three equations as,

$$\begin{gathered} x_1\left(i+1\right)=-0.25x_2\left(i\right)-0.125x_3\left(i\right)+0.375 \\ {x}_2\left(i+1\right)=0.1666x_2\left(i\right)-0.2487x_3\left(i\right)+0.4141 \\ {x}_3\left(i+1\right)=0.00595x_2\left(i\right)+0.09817x_3\left(i\right)-0.05652 \end{gathered}$$

By using the initial conditions,

For$x_1\left(1\right)$

$$\begin{gathered} {\mathrm{x}}_1\left(0+1\right)=-0.25{\mathrm{x}}_2\left(0\right)-0.125{\mathrm{x}}_3\left(0\right)+0.375 \\ {\mathrm{x}}_1\left(1\right)=0.25{\mathrm{x}}_2\left(0\right)-0.125{\mathrm{x}}_3\left(0\right)+0.375 \\ {\mathrm{x}}_1\left(1\right)=0.25\left(0\right)-0.125\left(0\right)+0.375 \\ {\mathrm{x}}_1\left(1\right)=0.375 \end{gathered}$$

For$x_2\left(1\right)$

$$\begin{gathered} {\mathrm{x}}_2\left(0+1\right)=-0.1666{\mathrm{x}}_2\left(0\right)-0.2487{\mathrm{x}}_3\left(0\right)+0.4141 \\ {\mathrm{x}}_2\left(1\right)=0.1666{\mathrm{x}}_2\left(0\right)-0.2487{\mathrm{x}}_3\left(0\right)+0.4141 \\ {\mathrm{x}}_2\left(1\right)=0.1666\left(0\right)-0.2487\left(0\right)+0.4141 \\ {\mathrm{x}}_2\left(1\right)=0.4141 \end{gathered}$$

For$x_3\left(1\right)$

$$\begin{gathered} {\mathrm{x}}_3\left(0+1\right)=-0.00595{\mathrm{x}}_2\left(0\right)+0.09817{\mathrm{x}}_3\left(0\right)-0.05652 \\ {\mathrm{x}}_3\left(1\right)=0.00595{\mathrm{x}}_2\left(0\right)+0.09817{\mathrm{x}}_3\left(0\right)-0.05652 \\ {\mathrm{x}}_3\left(1\right)=0.00595\left(0\right)+0.09817\left(0\right)-0.05652 \\ {\mathrm{x}}_3\left(1\right)=0.05652 \end{gathered}$$

Repeat the above steps to calculate the values for$x_1\left(n\right),x_2\left(n\right)and{x}_3\left(n\right)$.

The table of the iteration is shown below.

Hence the solution of the system is ${\mathrm{x}}_1=0.222803,{\mathrm{x}}_2=0.533775$and

$x_3=-0.058138$.

The Gauss-Seidel method converges more rapidly than Jacobean method.