... 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)
As I described
x1 + x2 > 2
x1 + 2x2 > 3
is expressed as AX > B, where A = [ 1 1 ; 1 2 ]; B = [2 3]T
Maybe it is a formal notation but I just would like to solve the problem.
x1 + 2x2 > 3
is expressed as AX > B, where A = [ 1 1 ; 1 2 ]; B = [2 3]T
Maybe it is a formal notation but I just would like to solve the problem.
Yeah!
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