Page 60 - Determinants and Their Applications in Mathematical Physics
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3.6 The Jacobi Identity and Variants  45
               (n+1)        (n)
          But A         = A   . Hence, A n P = 0. The result follows.
               i,n+1;j,n+1  ij
            Three particular cases of (B) are required for the proof of the next
          theorem.
            Put (i, p, q)=(r, r, n), (n − 1,r,n), (n, r, n) in turn:
                             (n)

                                     (n)        (n+1)
                            A rr   A rn     − A n A    =0,            (B 1 )
                            (n+1)
                          A       A
                                    (n+1)       r,n+1;rn
                            n+1,r   n+1,n

                           (n)     (n)
                          A      A             (n+1)

                           n−1,r   n−1,n     − A n A    =0,           (B 2 )
                           (n+1)   (n+1)       n−1,n+1;rn
                         A       A

                           n+1,r   n+1,n
                             (n)

                                     (n)        (n+1)
                                          − A n A      =0.            (B 3 )
                           A nr
                                   A nn
                            (n+1)
                            A     A  (n+1)      n,n+1;rn
                            n+1,r   n+1,n
          Theorem 3.7.
                             (n+1)        (n)     (n)
                             A
                             r,n+1;rn   A rr    A rn
                            (n+1)       (n)     (n)

                            A          A       A       =0.
                            n−1,n+1;rn  n−1,r   n−1,n
                             (n+1)        (n)
                            A
                                                  (n)
                             n,n+1;rn   A nr    A nn
          Proof. Denote the determinant by Q. Then,

                                       (n)     (n)
                                       A      A

                                       n−1,r   n−1,n
                               Q 11 =    (n)    (n)
                                        A nr   A nn
                                         (n)
                                   = A n A
                                         n−1,n;rn
                                         (n−1)
                                   = A n A   ,
                                         n−1,r
                               Q 21 = −A n A (n−1) ,
                                          rr
                                         (n)
                               Q 31 = A n A    .                    (3.6.17)
                                         r,n−1;rn
          Hence, expanding Q by the elements in column 1 and applying (B 1 )–(B 3 ),
                          (n+1)  (n−1)    (n+1)     (n−1)
                Q = A n A       A     − A          A
                         r,n+1;rn  n−1,r  n−1,n+1;rn  rr
                        (n+1)   (n)
                     + A       A                                    (3.6.18)
                        n,n+1;rn  r,n−1;rn

                                                      (n)     (n)
                              (n)
                      (n−1)           (n)      (n−1)   A    A
                  = A         A rr   A rn     − A     n−1,r   n−1,n
                             (n+1)
                                                      (n+1)
                                                              (n+1)
                            A       A  (n+1)    rr    A     A
                      n−1,r
                             n+1,r   n+1,n
                                                      n+1,r   n+1,n
                                  (n)

                        (n)               (n)
                     + A          A nr   A nn
                                 (n+1)
                                A       A  (n+1)
                        r,n−1;rn
                                 n+1,r   n+1,n

                                             (n−1)

                      (n+1)  (n)  (n)                 (n)
                  = A       A   A       −     A rr  A rr
                                             (n−1)
                                                     (n)
                                           A       A
                             nr
                      n+1,n      r,n−1;rn
                                             n−1,r   n−1,r
                                               (n−1)

                        (n+1)       (n)                 (n)
                     − A      A n−1 A      −     A rr  A rn     .   (3.6.19)
                                               (n−1)
                                                       (n)
                                              A       A
                        n+1,r       r,n−1;rn
                                               n−1,r   n−1,n
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