\[ \begin{bmatrix} 1.0 & 0.5 \\ 0.4 & 1.0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]
When's the best time to go to the dentist?
Tooth-hurtie!
Use Jacobi iteration to solve \( A \mathbf{x} = \mathbf{b} \) where
\[ A = \begin{bmatrix} 1.0 & 0.2 \\ 0.3 & 1.0 \end{bmatrix}, \,\,\,\, \mathbf{b} = \begin{bmatrix} 3 \\ 4 \end{bmatrix} \]
Since \( A \) is diagonally dominant, we iterate as follows:
\[ \mathbf{x}^{(k+1)} = \mathbf{b} + C\mathbf{x}^{(k)} \]
where
\[ C = I - A = \begin{bmatrix} 0 & -0.2 \\ -0.3 & 0 \end{bmatrix} \]
See Notes for the Octave program and results.
\[ \mathbf{x}^{(0)} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
Compare results of iteration to \( \mathbf{u} = A^{-1} \mathbf{b} \).
There will be some error between early iterates \( \mathbf{x}^{(k)} \) and \( \mathbf{u} \).
Measure \( \mathbf{e}^{(k)} = \mathbf{u}- \mathbf{x}^{(k)} \) using Frobenius vector norm.