Do not use det to try to determine if a matrix is singular or near-singular!
In this example, A is a nonzero scalar multiple of the identity and as such is most definitely NOT singular. But dA is 0. Why? Underflow. The condition number, on the other hand, correctly identifies this matrix as well conditioned.
A = 0.1*eye(350);
dA = det(A)
cA = cond(A)
In this example the first and second rows of B are almost the same, and dB reflects this fact. But the magnitude of the elements in C (a nonzero scalar multiple of B) makes the determinant reasonably large so C appears to be nonsingular. But the condition numbers cB and cC tell the tale.
B = [1 1; 1 1+eps];
C = flintmax * B;
dB = det(B)
dC = det(C)
cB = cond(B)
cC = cond(C)
