Gauss Seidel method in matlab to find the roots

Hello everyone, i have a porblem with Gauss Seidel.I have 4 equations and 4 unknown. I am giving the the initial conditions and ı also arranged the equations with respect to diagonal row ( so the diagonal side has to be bigger). If the maximum error will be lower than 1% error, ıt should give me values (x1,x2,x3,x4). May you please find where i am doing wrong. I really appreciate...

clc;clear; format ('long' , 'g'); x1(1)=1;x2(1)=1; x3(1)=1;x4(1)=1; error=1;

while error >=0,01

i=1;

    x1(i+1)=(-23+x2(i)-x3(i)+2*x4(i))/4;
    x3(i+1)=((2*x1(i+1)-6*x2(i)+3*x4(i)+21));
    x2(i+1)=(-x1(i+1)+11+5*x3(i+1)+x4(i))/2;
    x4(i+1)=(22+x1(i+1)-2*x2(i+1)+3*x3(i+1))/6;

     error_x1(i+1)=abs((x1(i+1)-x1(i)))/x1(i+1);
     error_x3(i+1)=abs((x3(i+1)-x3(i)))/x3(i+1);
     error_x2(i+1)=abs((x2(i+1)-x2(i)))/x2(i+1);
     error_x4(i+1)=abs((x4(i+1)-x4(i)))/x4(i+1);

    A=[error_x1 ;error_x3 ;error_x2;error_x4];
    error=max(A);

     i=i+1; 
  end
disp ( '                     x1(i+1)                      error(%) ')
disp([                       x1                        ' error_x1'])
disp ( '                     x3(i+1)                     error(%)')
disp([                       x3                        'error_x3'])
disp ( '                     x2(i+1)                       error(%)')
disp([                       x2                        'error_x2'])
disp ( '                     x4(i+1)                      error(%)')
disp([                       x4                         'error_x4'])

Answer 1

[Paul]

First of all 0,01 should be 0.01. Secondly, i=1; should be outside the while loop, else the value of i is reset every loop. I also fixed your display commands:

clc;clear; format ('long' , 'g'); x1(1)=1;x2(1)=1; x3(1)=1;x4(1)=1; error=1;
i=1;
while error >=0.01    
    x1(i+1)=(-23+x2(i)-x3(i)+2*x4(i))/4;
    x3(i+1)=((2*x1(i+1)-6*x2(i)+3*x4(i)+21));
    x2(i+1)=(-x1(i+1)+11+5*x3(i+1)+x4(i))/2;
    x4(i+1)=(22+x1(i+1)-2*x2(i+1)+3*x3(i+1))/6;
     error_x1(i+1)=abs((x1(i+1)-x1(i)))/x1(i+1);
     error_x3(i+1)=abs((x3(i+1)-x3(i)))/x3(i+1);
     error_x2(i+1)=abs((x2(i+1)-x2(i)))/x2(i+1);
     error_x4(i+1)=abs((x4(i+1)-x4(i)))/x4(i+1);
    A=[error_x1 ;error_x3 ;error_x2;error_x4];
    error=max(A);
     i=i+1; 
end
disp ( '                     x1(i+1)                      error(%) ')
disp([x1(end)  error_x1(end)])
disp ( '                     x3(i+1)                     error(%)')
disp([x2(end)  error_x2(end)])
disp ( '                     x2(i+1)                       error(%)')
disp([x3(end)  error_x3(end)])
disp ( '                     x4(i+1)                      error(%)')
disp([x4(end)  error_x4(end)])

Answer 2

[Ilke]

Firstly, i am really thankful for your answering. When i run your codes, it worked but it gave me wrong results and error is really high.True results must be [-4 3 -2 1].I think , the problem is with errors. i am looking where it is wrong. x1(i+1) error(%) -5.25 -1.19047619047619

                     x3(i+1)               error(%)
                    27.375         0.963470319634703

                     x2(i+1)                error(%)
                       7.5         0.866666666666667

                     x4(i+1)               error(%)
         -2.58333333333333         -1.38709677419355

[Paul]

A=[error_x1 ;error_x3 ;error_x2;error_x4];

should be:

A=[error_x1(i+1) ;error_x3(i+1) ;error_x2(i+1);error_x4(i+1)];

Still the values do not converge so you probably have some errors in your equations.