Gaussian elimination with backward substitution

 MATLAB Program:

 % Gaussian elimination with backward substitution

 n=input('Enter number of equations, n:  ');

 A = zeros(n,n+1);

 x = zeros(1,n);

 A=[4 2 3 8; 3 -5 2 -14; -2 3 8 27];

 nn = n-1; m = n+1;

 ichg = 0;

 i = 1;

 while i <= nn

    ip = i;

    while abs(A(ip,i)) <= 1.0e-20 & ip <= n

       ip = ip+1;

    end

    if ip == m

        fprinff('Method fails');

    else

        if ip ~= i

           for jj = 1:m

              c = A(i,jj);

              A(i,jj) = A(ip,jj);

              A(ip,jj) = c;

            end

            ichg = ichg+1;

        end

        jj = i+1;

        for j = jj:n

             xm = A(j,i)/A(i,i);

             for k = jj:m

                  A(j,k) = A(j,k) - xm * A(i,k);

             end

             A(j,i) = 0;

        end

     end

     i = i+1;

 end

 if abs(A(n,n)) <= 1.0e-20

    fprinff('Method fails \n');

 else

     x(n) = A(n,m) / A(n,n);

     for k = 1:nn

        i = nn-k+1;

        jj = i+1;

        sum = 0;

        for kk = jj:n

            sum = sum - A(i,kk) * x(kk);

        end

        x(i) = (A(i,m)+sum) / A(i,i);

     end

     fprintf('\n\nThe reduced system is: \n');

     for i = 1:n

        for j = 1:m

            fprintf(' %11.8f', A(i,j));

        end

        fprintf('\n');

     end

     fprintf('\nThe system has the solution:\n');

     for i = 1:n

         fprintf('           %11.8f  \n', x(i));

     end

 end

%Gaussian_Elimination_final.m

%Displaying Gaussian_Elimination_final.m.

Output:

Enter number of equations, n:  3

The reduced system is: 

  4.00000000  2.00000000  3.00000000  8.00000000

  0.00000000 -6.50000000 -0.25000000 -20.00000000

  0.00000000  0.00000000  9.34615385 18.69230769

The system has the solution:

           -1.00000000  

            3.00000000  

            2.00000000