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4.1 Alternants  55

          Hence,
                                 [x ij ] n [F ji ] n = I,
                                      [F ji ] n =[x ij ] −1
                                           =[X ] n .
                                               ji
                                               n
          The theorem follows.
          Theorem 4.2.
                               (n)               (n−1)
                             X    =(−1) n−j X n−1 σ   .
                               nj                n−j
          Proof. Referring to equations (A.7.1) and (A.7.3) in Appendix A.7,
                                          n−1

                                             (x n − x r )
                               X n = X n−1
                                          r=1
                                  = X n−1 f n−1 (x n )
                                  = X n−1 g nn (x n ).
          From Theorem 4.1,
                                                (n)
                                    (−1) n−j X n σ
                               (n)
                             X    =             n,n−j
                                         g nn (x n )
                               nj
                                                 (n)
                                  =(−1) n−j X n−1 σ   .
                                                 n,n−j
          The proof is completed using equation (A.7.4) in Appendix A.7.
          4.1.4  A Hybrid Determinant
          Let Y n be a second Vandermondian defined as
                                           j−1
                                    Y n = |y
                                           i  | n
          and let H rs denote the hybrid determinant formed by replacing the rth
          row of X n by the sth row of Y n .
          Theorem 4.3.
                                          g nr (y s )
                                       =        .
                                   H rs
                                          g nr (x r )
                                    X n
          Proof.
                        n
                           j−1
                     =    y   X  rj
                 H rs
                           s    n
                       j=1
                 X n
                          1              (n)
                               n
                     =           (−1) n−j σ   y j−1   (Put j = n − k)
                       g nr (x r )       r,n−j s
                              j=1
                              n−1
                          1
                     =           (−1) σ   y     .
                                     k (n) n−1−k
                       g nr (x r )     rk  s
                              k=0
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