Matlab Code for a rectangular matrix in row echelon form?

A=[1,2,1,3,3;2,4,0,4,4;1,2,3,5,5;2,4,0,4,7]

[n1,n2]=size(A); % # of rows and columns
p = (1:n1)';  % # each row in the matrix to pivot if necessary

for k = 1:n1-1 % Repeat the loop based on 1 - #rows
  [r,m] = max(abs(A(k:n1,k))) % r is the value of the max in the kth col and m is the     
                            % matrix location
  m = m+k-1 %Previous max matrix location plus the next 'k' step in the matrix

    if r~=0 %Checks to see if the col is zero
      break;
    end
  end

if (m~=k) %Checks to see if the max vector location equals the the next location    
          %step in the matrix. If not the switch the rows. 
  A([k m],:) = A([m,k],:)
  p([k m]) = p([m k])
end

i = k+1:n1;  %Count for the next row 
C(i,k) = A(i,k)/A(k,k) %Storing the multipliers for the kth col in Matrix C

j = k:n2 %Counting columns
A(i,j) = A(i,j)-C(i,k)*A(k,j) %Using the multiplier to complete forward elimination

Please see the matlab code that I wrote above, it did not complete the rectangular matrix in row echelon form. It only completed one cycle, I am having difficulty telling it to skip the zero column and check for the max in the adjacent column. It seems to stop when it gets to the zero column and not move on to check the next column. Can you please help?

Your help is greatly appreciated.

Thank you,

Addanis

[John D'Errico]

Please learn to format your code so it is readable. Read this .

Help those whom you would ask to help you. Make their job easier, and you will get help more quickly.

[Star Strider]

The rref function will do this for you.

[John D'Errico]

I somehow doubt that rref would be an acceptable solution to a homework problem. :)

At the same time, Paul DID post a not terrible attempt at code. It even seems to have comments in it. If I could read it, I'd be willing to take a look for the problem, after breakfast.

[John D'Errico]

I've edited your code, making it readable. Please see that it can now be read as code. Done once for you.

Answer 1

[Rahul Mishra]

m=input('no. of rows');
n=input('no. of columns');
for i=1:m
    for j=1:n
        A(i,j)=input('enter the entry');
    end
end
for z=1:m-1
    if(A(z,z)~=0)
        a=A(z,z)
        for k=z+1:m
            b=A(k,z)
            for l=1:n
                A(k,l)=a*A(k,l)-b*A(z,l)
            end
        end
    end
end
disp(A)