MATLAB code to find inverse of a Matrix by Gauss Emilination

  %% Write a program which can calculate inverse of a given matrix by

  %% by Gauss Elimination Method using complete pivot.  

    M = [1 2 13;3 24 5;25 65 7]; % Given matrix whose inverse is to be found

    A = M;

    Ni = length(A(1,:)); % number of rows of given matrix

    Nj =length(A(:,1));   

    if abs(det(A))<0.00000000001

        fprintf('The matrix determinant is zero, the result found may not be correct');

    end     

    px =1:Ni;

    qy =1:Ni;  
 

    %%main loop; find pivot entry in columns qy[k], qy[k+1], ...,qy[n-1]   

    %% and in rows px[k], px[k+1],.....px[n-1]

  

    for k = 1: Ni-1

        pct = k; qdt = k;

        aet =0;

        for i=k:Ni

            for j=k:Ni

                tmp = abs(A(uint8(px(i)),uint8 (qy(j))));

                if (tmp>aet)

                    aet = tmp;  pct = i;qdt = j;

                end

            end

        end

      

        if ~aet

            fprintf('pivot is zero, No inverse possible \n');

        end

     

     % Swap the rows and columns to get highest number in the diagonal

        px([k pct]) = px([pct k]);%//swap px[k] and px[pct]                        

        qy([k qdt]) = qy([qdt k]) ; 

        %% eliminate column entries (make them zero) logically below and right to the entry mx[px[k]][qy[k]]

        for i=k+1:Ni   

            if A(px(i),qy(k)) ~= 0  % if the entry is not equal to zero        

                mult = A(px(i),qy(k))/A(px(k),qy(k)); % get the multiplying factor

                A(px(i),qy(k))= mult;

                for j=k+1:Ni

                    A(px(i),qy(j)) = A(px(i),qy(j)) - mult*A(px(k),qy(j)); % entry will become zero

                end

            end

        end

    end

  

    AI = eye(size(A));  %% Identiy Matrix

    %%forward substitution 

    for k=1:Ni

        for i =2:Ni

            for j =1:i-1

                AI(px(i),k) = AI(px(i),k) - A(px(i),qy(j))*AI(px(j),k);

            end

        end

    end      

    InMat = zeros(Ni,Nj); %InMat will be the required inverse after back substitution.  

    %% backward substitution

    for k=1:Ni

        for i = Ni:-1:1

            for j = i+1:Ni

                AI(px(i),k) = AI(px(i),k) - A(px(i),qy(j))* InMat(qy(j),k);

            end

        InMat(qy(i),k) = AI(px(i),k) / A(px(i),qy(i));

        end

    end

    %% wee can also find error of our result by subtracting the identity matrix

    %% from the product of given matrix (M) and our result (InMat)

    I = eye(Ni,Nj);

    error = InMat*M - I;

   %% output of the results

    fprintf('The given matrix is \n\n');

    fprintf(' %f %f %f \n',M);

    fprintf('The inverse of the matrix is \n\n');

    fprintf(' %f  %f  %f \n',InMat);

    fprintf('The error in the calculate of the invese of matrix is \n\n');

    fprintf('\n %f  %f  %f \n',error);