How can I solve this following problem? (overcome * mark)

My code is :

%Start
clear all
clc 
%Input required values
fprintf('***********************Gauss Jodern Elimination Method*************************\n')
fprintf('***************<strong>Code has been created by Pulak(181201)</strong>***********************\n')
format rational
n=input('How many variables?:  ');
A=input('coefficient matrix: ');
for i=1:n
    for j=1:n
        a(i,j)=A(i,j);
       
    end 
end
AB=a;
B=input('Constant matrix: ');
for i=1:n
   b(i)=B(i);
   
end
disp('The given equations in matrix form is or Augmented Matrix:')
AB=[A B]
%checking the ranks

%Finding determinant
dett = det(a);
%Check Elemination
if dett == 0
    print('This system unsolvable because det(C) = 0 ')
    return
else
     if rank(A)==rank(AB)
         disp('Unique solution exit')
     else
         disp('Unique solution doesnot exit')
     end 
 
    j=1;
    for i = 1:n-1 
           if (i==j)
                if (AB(i,j)==0) 
                    for q=1:n-1
                        if (AB(i+q,j)~=0)
                          c = AB(i,:); 
                    AB(i,:)= AB(i+q,:); 
                    AB(i+q,:) = c;  
                    break
                        end  
                    end
                   
                end
                AB(i,:)= (1/AB(i,j))* AB(i,:);
                for k=i+1:n
                    AB(k,:)=(-AB(k,j)* AB(i,:)) + AB(k,:);
                    
                end
           end
        j=j+1;        
    end
 AB(n,:)= (1/AB(n,n))* AB(n,:);

for j=n:-1:2
    for i=j-1:-1:1
        AB(i,:)=AB(i,:)-AB(j,:)*(AB(i,j))
    end
end
disp('The final Matrix is:')
AB
for s=1:n
    x(s)=AB(s,n+1)
end

  for i=1:n
       ans=x(i);
       fprintf('\nThe %d-th solution is x(%d)= %.6f',i,i,ans)
   end
 fprintf('\n\n')   
 fprintf('***********************<strong>Code has been created by Pulak(181201)</strong>***********************\n')

end

but in the ouput:

***********************Gauss Jodern Elimination Method*************************

***************Code has been created by Pulak(181201)***********************

How many variables?: 4

coefficient matrix: [1 -0.7 0.3 0.5;-6 8 -1 -4;3 1 4 11;5 -9 -2 4]

Constant matrix: [0.2;5;2;7]

The given equations in matrix form is or Augmented Matrix:

AB =

1 -7/10 3/10 1/2 1/5

-6 8 -1 -4 5

3 1 4 11 2

5 -9 -2 4 7

Unique solution exit

The final Matrix is:

AB =

1 * 0 * 3531/923

0 1 0 * 1091/333

0 * 1 * -2063/324

0 * 0 1 1067/923

The 1-th solution is x(1)= 3.825569

The 2-th solution is x(2)= 3.276273

The 3-th solution is x(3)= -6.367281

The 4-th solution is x(4)= 1.156013

***********************Code has been created by Pulak(181201)***********************

My question is why * is seen instead of 0? And how can I overcome this problem?

Answer 1

[Steven Lord]

From the help text for the format function:

format RAT Approximation by ratio of small integers. Numbers

with a large numerator or large denominator are

replaced by *.

Most likely this means those elements are small (but non-zero) numbers that have large denominators.

format rational
x = [1, 1e-10]

x = 
       1              *       

[PULAK Kumer]

How can i solve this problem sir?

[Steven Lord]

Some options:

Replace those small values with 0.

Replace them with close approximations that have small denominators.

Display your data using a different display format.