Re: Implementing a Jacobi iterative method for Ax=b

[Ozzy Lash]

Oh, and if you permute A, I think you will also need to permute b so
that they match, and then permute the result back to the original
form.  I didn't try a matrix with any 0s on the diagonal. I used:

A=[-500, 2, 3; 3, 200, 1; 4,5,600]
b=[1;2;3]