Because the functions 
, 
, and 1 are linearly dependent, we should find that the following matrix is somewhat ill-conditioned.
, , and 1 are linearly dependent, we should find that the following matrix is somewhat ill-conditioned.t = linspace(0,3,400)';
A = [ sin(t).^2, cos((1+1e-7)*t).^2, t.^0 ];
kappa = cond(A)
Now we set up an artificial linear least squares problem with a known exact solution that actually makes the residual zero.
x = [1;2;1];
b = A*x;
Using backslash to find the solution, we get a relative error that is about κ times machine epsilon.
x_BS = A\b;
observed_err = norm(x_BS-x)/norm(x)
max_err = kappa*eps
If we formulate and solve via the normal equations, we get a much larger relative error. With 
, we may not be left with more than about 2 accurate digits.
, we may not be left with more than about 2 accurate digits.N = A'*A;
x_NE = N\(A'*b);
observed_err = norm(x_NE-x)/norm(x)
digits = -log10(observed_err)