Jacobi Method - Example

Example

A linear system of the form with initial estimate is given by

A= \begin{bmatrix} 2 & 1 \\ 5 & 7 \\ \end{bmatrix}, \ b= \begin{bmatrix} 11 \\ 13 \\ \end{bmatrix} \quad \text{and} \quad x^{(0)} = \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} .

We use the equation, described above, to estimate . First, we rewrite the equation in a more convenient form, where and . Note that where and are the strictly lower and upper parts of . From the known values

D^{-1}= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix}, \ L= \begin{bmatrix} 0 & 0 \\ 5 & 0 \\ \end{bmatrix} \quad \text{and} \quad U = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} .

we determine as

T= \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix} \left\{ \begin{bmatrix} 0 & 0 \\ -5 & 0 \\ \end{bmatrix} + \begin{bmatrix} 0 & -1 \\ 0 & 0 \\ \end{bmatrix}\right\} = \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} .

Further, C is found as

C = \begin{bmatrix} 1/2 & 0 \\ 0 & 1/7 \\ \end{bmatrix} \begin{bmatrix} 11 \\ 13 \\ \end{bmatrix} = \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix}.

With T and C calculated, we estimate as :

x^{(1)}= \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} + \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix} = \begin{bmatrix} 5.0 \\ 8/7 \\ \end{bmatrix} \approx \begin{bmatrix} 5 \\ 1.143 \\ \end{bmatrix} .

The next iteration yields

x^{(2)}= \begin{bmatrix} 0 & -1/2 \\ -5/7 & 0 \\ \end{bmatrix} \begin{bmatrix} 5.0 \\ 8/7 \\ \end{bmatrix} + \begin{bmatrix} 11/2 \\ 13/7 \\ \end{bmatrix} = \begin{bmatrix} 69/14 \\ -12/7 \\ \end{bmatrix} \approx \begin{bmatrix} 4.929 \\ -1.713 \\ \end{bmatrix} .

This process is repeated until convergence (i.e., until is small). The solution after 25 iterations is

x=\begin{bmatrix} 7.111\\ -3.222 \end{bmatrix} .

Read more about this topic:  Jacobi Method

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