Gauss elimination and Gauss Jordan methods using MATLAB code

gauss.m

% Code from "Gauss elimination and Gauss Jordan methods using MATLAB"
% https://www.youtube.com/watch?v=kMApKEKisKE

a = [3 4 -2 2 2
    4 9 -3 5 8
    -2 -3 7 6 10
    1 4 6 7 2];


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Gauss elimination method [m,n)=size(a);
[m,n]=size(a);
for j=1:m-1
    for z=2:m
        if a(j,j)==0
            t=a(j,:);a(j,:)=a(z,:);
            a(z,:)=t;
        end
    end
    for i=j+1:m
        a(i,:)=a(i,:)-a(j,:)*(a(i,j)/a(j,j));
    end
end
x=zeros(1,m);
for s=m:-1:1
    c=0;
    for k=2:m
        c=c+a(s,k)*x(k);
    end
    x(s)=(a(s,n)-c)/a(s,s);
end
disp('Gauss elimination method:');
a
x'
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Gauss-Jordan method
[m,n]=size(a);
for j=1:m-1
    for z=2:m
        if a(j,j)==0
            t=a(1,:);a(1,:)=a(z,:);
            a(z,:)=t;
        end
    end
    for i=j+1:m
        a(i,:)=a(i,:)-a(j,:)*(a(i,j)/a(j,j));
    end
end

for j=m:-1:2
    for i=j-1:-1:1
        a(i,:)=a(i,:)-a(j,:)*(a(i,j)/a(j,j));
    end
end

for s=1:m
    a(s,:)=a(s,:)/a(s,s);
    x(s)=a(s,n);
end
disp('Gauss-Jordan method:');
a
x'

what is the difference bw gauss Jordan method and gauss Jordan elimination

in line 14 , it will be
for z=j+1:m
otherwise
for z=2:m
will not work for
a=[2 1 -1 2 5
4 5 -3 6 9
4 2 -2 9 8
4 11 -4 8 2];

a(j,:) means?

Can i get the matlab gui implementation of gauss elimination ?

Good job

if your matrix is changed as shown below, does your program work?

a = [3 4 -2 2 2
4 0 -3 5 8
-2 -3 0 6 10
1 4 6 7 2];
thanks

Just as functoin:

function [x] = gauss(a,y)
%GAUSS Summary of this function goes here
%   Detailed explanation goes here
[m,n]=size(a);
a(:, n+1) = y;
[m,n]=size(a);
for j=1:m-1
    for z=2:m
        if a(j,j)==0
            t=a(1,:);a(1,:)=a(z,:);
            a(z,:)=t;
        end
    end
    for i=j+1:m
        a(i,:)=a(i,:)-a(j,:)*(a(i,j)/a(j,j));
    end
end

for j=m:-1:2
    for i=j-1:-1:1
        a(i,:)=a(i,:)-a(j,:)*(a(i,j)/a(j,j));
    end
end

for s=1:m
    a(s,:)=a(s,:)/a(s,s);
    x(s)=a(s,n);
end

end

Can we find inverse using this method ? What is the final output which we get ?

good job

if your matrix is changed as shown below, your program does work!

a=[0 0 0 5 ; 4 0 2 5; 1 3 0 2; 3 4 2 0]

y=[16;10;13;-1]
thanks