“…Situation Two: A 1;1 < 0, A 2;2 < 0, A 3;3 ¼ 0: By equations ( 6), (10), and ( 14), we get A 1;2 þ A 2;1 < 0, A 3;2 þ A 2;3 < 0, A 1;3 þ A 3;1 < 0. By equation (7), we have A 1;2 ¼ A 1;3 þ A 3;2 . By equation (8), we obtain A 1;3 ¼ maxfA 1;2 þ A 2;3 , A 1;3 g, by equation ( 9) we know A 2;1 ¼ A 2;3 þ A 3;1 , by equation (11) we have A 2;3 ¼ maxfA 2;1 þ A 1;3 , A 2;3 g, by equation ( 12) we get A 3;1 ¼ maxfA 3;2 þ A 2;1 , A 3;1 g, and by equation ( 13) we obtain A 3;2 ¼ maxfA 3;1 þ A 1;2 , A 3;2 g.…”