MATLAB has a function cond to compute 
. The family of Hilbert matrices is famously badly conditioned. Here is the 
case.
. The family of Hilbert matrices is famously badly conditioned. Here is the case. A = hilb(7);
kappa = cond(A)
Next we engineer a linear system problem to which we know the exact answer.
x_exact = (1:7)';
b = A*x_exact;
Now we perturb the data randomly but with norm 
.
.randn('state',333); % reproducible results
dA = randn(size(A)); dA = 1e-12*(dA/norm(dA));
db = randn(size(b)); db = 1e-12*(db/norm(db));
We solve the perturbed problem using built-in pivoted LU and see how the solution was changed.
x = (A+dA) \ (b+db);
dx = x - x_exact;
Here is the relative error in the solution.
rel_error = norm(dx) / norm(x_exact)
And here are upper bounds predicted using the condition number of the original matrix.
b_bound = kappa * 1e-12/norm(b)
A_bound = kappa * 1e-12/norm(A)
Even if we don't make any manual perturbations to the data, machine epsilon does when we solve the linear system numerically.
x = A\b;
rel_error = norm(x - x_exact) / norm(x_exact)
rounding_bound = kappa*eps
Because 
, it's possible to lose 8 digits of accuracy in the process of passing from 
and 
to 
. That's independent of the algorithm; it's inevitable once the data are expressed in double precision.
, it's possible to lose 8 digits of accuracy in the process of passing from and to . That's independent of the algorithm; it's inevitable once the data are expressed in double precision. Now we choose an even more poorly conditioned matrix from this family.
A = hilb(14);
kappa = cond(A)
Before we compute the solution, note that κ exceeds 1/eps. In principle we might end up with an answer that is completely wrong.
rounding_bound = kappa*eps
MATLAB will notice the large condition number and warn us not to expect much from the result.
x_exact = (1:14)';
b = A*x_exact; x = A\b;
In fact the error does exceed 100%.
relative_error = norm(x_exact - x) / norm(x_exact)