Hi Terri,
Looks like it is not possible. You have
[F1;F2;F3] = [10 2 10
15 20 8
6 4 12] * [x1;x2;x3]
Now if F1-F2 > 0 ident2
and F1-F3 > 0 ident3
then 2*F1-F2-F3 > 0 ident4 necessary but not sufficient
but that last quantity is [2 -1 -1]*[F1;F2;F3]
= [2 -1 -1]*[10 2 10
15 20 8
6 4 12] * [x1;x2;x3];
= [-1 -20 0] * [x1;x2;x3]
and that quantity can't be greater than zero. If you had allowed equality with the F's
F1-F2 >= 0 and F1-F3 >= 0
then x1 = x2 = 0, x3 = 1 might have been possible but it gives
which satisfies the equality-allowed version of ident4 but not the equality-allowed version of ident3. So it goes.
Best Answer