Page 81 - Determinants and Their Applications in Mathematical Physics
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66   4. Particular Determinants

                           = |c ij | n ,
          where
                     n
                           r+1
               c ij =  (−1)   a n+1−i,r a n+1−j,n+1−r  (put r = n +1 − s)
                    r=1
                            n
                                  s+1
                        n+1
                  =(−1)       (−1)   a n+1−j,s a n+1−i,n+1−s
                           s=1
                  =(−1) n+1 c ji .                                   (4.3.4)
          The theorem follows.

          Theorem 4.9. A skew-symmetric determinant of odd order is identically
          zero.
          Proof. Let A  ∗    denote the determinant obtained from A 2n−1 by
                        2n−1
          changing the sign of every element. Then, since the number of rows and
          columns is odd,
                                  A ∗   = −A 2n−1 .
                                    2n−1
          But,

                               A ∗   = A T   = A 2n−1 .
                                2n−1    2n−1
          Hence,
                                     A 2n−1 =0,
          which proves the theorem.
                         (2n)
            The cofactor A   is also skew-symmetric of odd order. Hence,
                         ii
                                       (2n)
                                      A    =0.                       (4.3.5)
                                       ii
          By similar arguments,
                                     (2n)     (2n)
                                    A    = −A    ,
                                     ji       ij
                                   (2n−1)   (2n−1)
                                  A      = A      .                  (4.3.6)
                                   ji       ij
          It may be verified by elementary methods that
                                 2
                           A 2 = a ,                                 (4.3.7)
                                 12
                                                       2
                           A 4 =(a 12 a 34 − a 13 a 24 + a 14 a 23 ) .  (4.3.8)
          Theorem 4.10.    A 2n is the square of a polynomial function of its
          elements.
          Proof. Use the method of induction. Applying the Jacobi identity
          (Section 3.6.1) to the zero determinant A 2n−1 ,
                                (2n−1)
                               A       A  (2n−1)


                                ii      ij      =0,
                                (2n−1)  (2n−1)
                                A      A
                                ji      jj
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