MATLAB: How to rearrange an out of order upper triangular matrix

MATLABmatrixmatrix manipulationsort

PREREQ —— A is a nxn matrix with random numbers, b is the corresponding answer vector,
here is my code:
function [A,b,x] = ordersolve(A,b)                      % I'm relatively new to matlab, so I wanted to put my thoughts on what I think my
                                                        % code is doing over here. If my logic doesn't match up to my code; which it
[n,m] = size(A);                                        % most likely doesn't considering this post... please tell me! I want to learn!
rowct = 1;                              % initialize row count to 1,
A2 = A;                                 % initialize a matrix of the same size, doesn't matter what it has in it because I'm replacing            
                                        % rows anyway, but it could have been made with zeros(n,m).
    for i = 1:1:m                       % for each column,
        for j = 1:1:n                       % for each row -- I want it to go through column by column :: for each row number in col m, check. 
            A1 = A(j,i)                     % create new number to hold the content of A(j,i).
            if(A1 ~= 0)                     % if A1 isnt =  to 0,
                
                A2(rowct,:) = A(j,:);       % A2 in the current row, starting at 1, is replaced with A row @ where number is not 0.
                A(j,:) = [];                % A then has its row erased so next time through the loop, that row won't be there anymore.
                
                b2(rowct,:) = b(j,:);       % ^ same logic
                b(j,:) = [];
                
                rowct = rowct + 1;          % increase row count by 1; only whenever we find a diagonalnum.
                
            end
            
        end
    end
    A = A2;     % replace variables that will be returned with the recalculated A2 and b2 matrices.
    b = b2;
     
% this is backward substitution                    % <<---------- This whole part below was given already; I assume it runs correctly exactly as presented.
x = zeros(n,1);
for i = n:-1:1
    
    x(i) = b(i);
    
    for j = i+1:n
        x(i) = x(i) - A(i,j)*x(j);
    end
    
    x(i) = x(i)/A(i,i);
end
end

Best Answer

function [A,b,x] = ordersolve(A,b)
[n,m] = size(A);
rowct = 1;
A2 = zeros(n,n);
x = 0;
    for i = 1:1:m
        c = n - x - 1
        for j = 1:1:c
            A1 = A(j,i)
            if(A1 ~= 0)
                
                A2(rowct,:) = A(j,:);
                A(j,:) = [];
                
                b2(rowct,:) = b(j,:);
                b(j,:) = [];
                
                rowct = rowct + 1;
                A2
                x = x-1
                break
                
            end
            
        end
    end
    A = A2;
    b = b2;
     
% this is backward substitution
x = zeros(n,1);
for i = n:-1:1
    
    x(i) = b(i);
    
    for j = i+1:n
        x(i) = x(i) - A(i,j)*x(j);
    end
    
    x(i) = x(i)/A(i,i);
end
end
figured it out!
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