Implementing Richardson's Iterative Method

I'm trying to implement Richardson's iterative method to solve Ax=b equation. I want to see the values in my matrix. But I wrote it in a way, that I don't know how to do it. I thought about writing it as three separate equations instead of vector form, but I'm not quite sure how you would do that. This is my current code:

format long
A = [9 1 1;
    2 10 3;
    3 4 11];
b = [10;
    19;
    0];
x = [0;
    0;
    0];
G=eye(3)-A; %I-A

z = [0,x'];
for k=1:30
  x = G*x + b;
  z = [k,x'];

end

fprintf('Number of Iterations: %d \n', k);
display(z);

Answer 1

[Geoff Hayes]

Chris - are you trying to see the evolution of x over all iterations? Is that why you have the

 z = [k, x'];

to give the x for the kth iteration? If so, then you could replace the above with

 z = [z ; [k x']];

so that we always concatenate the new values to the previous ones. Try the above and see what happens!

[cee878]

Yes, that's exactly what I wanted to do. But conceptually, could you tell me what that new vector z does. I really want to understand this. Thanks!

[cee878]

Also, could you tell me how to restrict the values to 6 decimal places, when I'm trying to print the values out?

[Geoff Hayes]

Chris - on each iteration of the loop, we add a row to z by concatenating the new row to the previous ones as

 z = [k, x'];

Use the debugger to step through the code and see this happen.

As for restricting the values to 6 decimal places, see fprintf and in particular the format specification. For a single floating point value, you would do something like

 fprintf('pi to six decimals is %.6f\n',pi);