600 则有P-1AP=030=A,又P1 3 003 12 A=PAP-I 600 10030 1-12 3 101八(003 12-1 A=(41,12P2,A3Py3n3) Solution 2. A(P1, P2, P3)=(M P1, n2 p2 K, 0 0 3 0 3 0 6 0 0 1 =            = − 则有P AP , 1 2 1 1 1 2 1 1 1 3 1 1           − − = − − − 又P −1  A = PP , 1 2 1 1 1 2 1 1 1 3 1 0 0 3 0 3 0 6 0 0 1 0 1 1 1 0 1 1 1           − −  − −                      − − = . 1 1 4 1 4 1 4 1 1           = Solution 2. ( , , ) ( , , ) A p1 p2 p3 = 1 p1 2 p2 3 p3 ( , , ) . 1 1 1 2 2 3 3 − A =  p  p  p P