... Can you prove that?Note what I said, for equation: if a = b then c = d a+b=c+d but for inequation, if a > b then c > d => a+b > c+d but a+b > c+d (not =>) c > d A simple counterexample: A = [ 1 1 ; 1 2] is PD B = [2;3] and AX > B (1) But we can't say X > A^-1 * b = [1;1] (2) Since obviously x1 = 0, x2 = 3 fits (1) but not (2)