Array indices must be positive integers or logical values..........??

What is wrong in following code??

%% Jacobi's method for the solution of system of linear equations
clc 
clear all
close all
format short
neq=input('Enter the number of equations i.e. neq:');
a=zeros(neq,neq);
for i=1:neq
    for j=1:neq
        a(i,j)=input('\nEnter the co-efficent values of the system :');
    end
end
for i=1:neq
    b(i)=input('\nEnter the constant values b: ');
end
fprintf('Iinitial guess values of the system\n');
x0=0.0; y0=0.0;  z0=0.0; 
itr=input('\nEnter the total number of iterations :');
for j=1:itr
    for i=0:neq
    	x(i+1)= (b(i)-a(i,i+1)*y0-a(i,i+2)*z0)/a(i,i);
        y(i+1)= (b(i+1)-a(i+1,i)*x0-a(i+1,i+2)*z0)/a(i+1,i+1);
        z(i+1)= (b(i+2)-a(i+2,i)*x0-a(i+2,i+1)*y0)/a(i+2,i+2);
        Ex=abs(x(i+1)-x(i));
        Ey=abs(y(i+1)-y(i));
        Ez=abs(z(i+1)-z(i));
        x0=x(i+1);
        y0=y(i+1);
        z0=z(i+1);
    end
end

Answer 1

[Walter Roberson]

for i=1:neq
  b(i)=input('\nEnter the constant values b: ')

You store into b(1), b(2) and so on all the way up to b(neq)

for i=0:neq
  x(i+1)= (b(i)-a(i,i+1)*y0-a(i,i+2)*z0)/a(i,i);

b(1) to b(neq) are stored into, but you want to retrieve b(i) with i starting from 0 so you want b(0) to b(neq). 0 is not a valid subscript.

You created a as neq by neq but here because of the a(i, i+1) you want to access a(0,1) to a(neq, neq+1)

[Dr. WAS]

Dear when I used your suggestion insteed of i=0 use i=1but response is same

for i=1:neq

x(i+1)= (b(i)-a(i,i+1)*y0-a(i,i+2)*z0)/a(i,i);

[Walter Roberson]

You are accessing up to a(i,i+2) in that code. That is only going to work if i+2 is no more than the number of columns in a . The number of columns in a is neq -- so that code can only work if the upper limit on i is no more than neq-2 .