I can see what you're going for, but you've got a couple issues.
The reason that you're not getting any pattern out is that the k=k+1 increment happens inside the conditional, where it should be happening in every iteration of the inner loop. That said, there's still a problem with using k like this. In this context, k is a linear index, so this solution will only produce a checkerboard if n is odd. otherwise, it'll produce stripes. The way you would fix this is to consider both i and j:
x = ones(n,n);
k = 0;
for i =1:n
for j = 1:n
if xor(rem(i, 2) == 1, rem(j, 2) == 1)
x(i,j) = 0;
end
end
end
a = x;
That said, you should start to see how xor and checkerboards go hand in hand. Ideally, you shouldn't even need any loops.
% assuming nxn square output
n=5;
x=mod(1:n,2);
a=bsxfun(@xor,x,~x')
or
% generalized mxn output
m=4;
n=5;
x=mod(1:n,2);
y=mod(1:m,2);
a=bsxfun(@xor,x,~y')
