Question

Express the following set of equations in the form of (6.2.6), and then solve using the Jacobi iterative method with$\mathbf{\epsilon }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$and withlocalid="1655268416489" ${\mathbf{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{x}}_{\mathbf{3}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

localid="1655268421093" $\mathbf{\left[}\begin{array}{ccc}\mathbf{10}& \mathbf{-}\mathbf{2}& \mathbf{-}\mathbf{4}\\ \mathbf{-}\mathbf{2}& \mathbf{6}& \mathbf{-}\mathbf{2}\\ \mathbf{-}\mathbf{4}& \mathbf{-}\mathbf{2}& \mathbf{10}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{c}{\mathbf{x}}_{\mathbf{1}}\\ {\mathbf{x}}_{\mathbf{2}}\\ {\mathbf{x}}_{\mathbf{3}}\end{array}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{-}\mathbf{2}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{3}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}$

Short answer

The solution of the system is$x_1=-0.1382,x_2=0.4336$and$x_3=-0.0668$.

Step-by-step solution

Write the given data from question.

The initial conditions are

$$\begin{gathered} x_1\left(0\right)=1 \\ {x}_2\left(0\right)=1 \\ {x}_3\left(0\right)=0, \end{gathered}$$

The matrix for the system,

$\left[\begin{array}{ccc}10& -2& -4\\ -2& 6& -2\\ -4& -2& 10\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right]=\left[\begin{array}{c}-2\\ 3\\ -1\end{array}\right]$

The specific tolerance level,$\epsilon =0.05.$

Determine the equation to find the solution of the system.

The general equation of the system is given as follows.

$\mathbf{Ax}\mathbf{=}\mathbf{y}$ …… (1)

Here$\mathbf{x}\mathbf{,}\mathbf{y}$are the N vectors A andis the$\mathbf{N}\mathbf{\times }\mathbf{N}$square matrix.

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

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

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

The equation for the Jacobi method as,

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

Calculate the solution for the system.

Compare the given system equation with equation (1)$A=\left[\begin{array}{ccc}10& -2& -4\\ -2& 6& -2\\ -4& -2& 10\end{array}\right],x=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right],y=\left[\begin{array}{c}-2\\ 3\\ 1\end{array}\right]$

For Jacobi, D consists of the diagonal elements of the A matrix.

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

Calculate the inverse of the D matrix.

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

Calculate the matrix M for Jacobi method,

$\mathrm{Substitude}\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{6}& 0\\ 0& 0& \frac{1}{10}\end{array}\right]\mathrm{for}{\mathrm{D}}^{-1}\left[\begin{array}{ccc}10& 0& 0\\ 0& 6& 0\\ 0& 0& 10\end{array}\right]\mathrm{for}\mathrm{D}\mathrm{and}\left[\begin{array}{ccc}10& -2& -4\\ -2& 6& -2\\ -4& -2& 10\end{array}\right]\mathrm{for}\mathrm{A}\mathrm{into}$

equation (2).

$$\begin{gathered} M=\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{6}& 0\\ 0& 0& \frac{1}{10}\end{array}\right]\left(\left[\begin{array}{ccc}10& 0& 0\\ 0& 6& 0\\ 0& 0& 10\end{array}\right]-\left[\begin{array}{ccc}10& -2& -4\\ -2& 6& -2\\ -4& -2& 10\end{array}\right]\right) \\ M=\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{6}& 0\\ 0& 0& \frac{1}{10}\end{array}\right]\left[\begin{array}{ccc}0& 2& 4\\ 2& 0& 2\\ 4& 2& 0\end{array}\right] \\ M=\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{6}& 0\\ 0& 0& \frac{1}{10}\end{array}\right] \end{gathered}$$

Expressed the equation of the system,

$\mathrm{Substitude}\left[\begin{array}{ccc}0& \frac{1}{5}& \frac{2}{5}\\ \frac{1}{3}& 0& \frac{1}{3}\\ \frac{2}{5}& \frac{1}{5}& 0\end{array}\right]\mathrm{for}M,\left[\begin{array}{cc}{x}_1& \left(i\right)\\ x_2& \left(i\right)\\ x_3& \left(i\right)\end{array}\right]\mathrm{for}x\left(i\right),\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{6}& 0\\ 0& 0& \frac{1}{10}\end{array}\right]\mathrm{for}{D}^{-1},and\left[\begin{array}{c}-2\\ 3\\ -1\end{array}\right]$

y into equation (3).

$\left[\begin{array}{cc}{x}_1& \left(i+1\right)\\ x_2& \left(i+1\right)\\ x_3& \left(i+1\right)\end{array}\right]=\left[\begin{array}{ccc}0& \frac{1}{5}& \frac{2}{5}\\ \frac{1}{3}& 0& \frac{1}{3}\\ \frac{2}{5}& \frac{1}{5}& 0\end{array}\right]\left[\begin{array}{cc}{x}_1& \left(i\right)\\ x_2& \left(i\right)\\ x_3& \left(i\right)\end{array}\right]+\left[\begin{array}{ccc}\frac{1}{10}& 0& 0\\ 0& \frac{1}{10}& 0\\ 0& 0& \frac{1}{10}\end{array}\right]\left[\begin{array}{c}-2\\ 3\\ -1\end{array}\right]$

By usingthe initial conditions,

For$x_1\left(1\right)$

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.=

The iterations are converging at the${10}^{th}$iteration,

For$x_1$.