# Getting back the solution to Ax=b after reordering A

4 views (last 30 days)
Jeff Chen on 19 Oct 2019
Edited: Bruno Luong on 21 Oct 2019
When I run the code below, x does not equal to A\b i.e. the ordering is wrong. It is my understanding that x_ordered(P_amd) should reorder the elements according to [2 3 4 5 1] - is that incorrect?
clc
clear
close all
A=[4 1 2 0.5 2;1 0.5 0 0 0;2 0 3 0 0;0.5 0 0 5/8 0;2 0 0 0 16];
% P_amd=amd(A);
P_amd=[2 3 4 5 1];
A_ordered=A(P_amd,P_amd);
b=[1 2 3 4 5]';
b_ordered=b(P_amd);
A\b
x_ordered=A_ordered\b_ordered
x=x_ordered(P_amd)

Bruno Luong on 20 Oct 2019
This is correct:
clc
clear
close all
A=[4 1 2 0.5 2;1 0.5 0 0 0;2 0 3 0 0;0.5 0 0 5/8 0;2 0 0 0 16];
% P_amd=amd(A);
P_amd=[2 3 4 5 1];
A_ordered=A(P_amd,P_amd);
b=[1 2 3 4 5]';
b_ordered=b(P_amd);
x=A\b;
x=x(P_amd)
x_ordered=A_ordered\b_ordered

Jeff Chen on 21 Oct 2019
Hi thanks for that,
A follow up question: I noticed you permuted x as well, but if Ax=b is the original system I want to solve, how would I go from x_ordered (solution to the reorderd system) to x, the solution to the original problem - this is the question I wanted to ask, sorry for the confusion.
Bruno Luong on 21 Oct 2019
Use the inverse of the permutation
clc
clear
close all
A=[4 1 2 0.5 2;1 0.5 0 0 0;2 0 3 0 0;0.5 0 0 5/8 0;2 0 0 0 16];
P_amd=[2 3 4 5 1];
A_ordered=A(P_amd,P_amd);
b=[1 2 3 4 5]';
b_ordered=b(P_amd);
x=A\b
x_ordered=A_ordered\b_ordered;
Pi_amd(P_amd) = 1:length(P_amd); % inverse of the permutation
x_ordered(Pi_amd)