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4.3 Skew-Symmetric Determinants  69

          and let B n+1 denote the skew-symmetric determinant obtained by
          bordering A n by the row


                               −1 − 1 − 1 ··· − 10
                                                  n+1
          below and by the column
                                               T
                                   111 ··· 10
                                              n+1
          on the right.

          Theorem 4.12 (Muir and Metzler).    B n+1 is expressible as a skew-
          symmetric determinant of order (n − 1).

          Proof. The row and column operations

                         R = R i + a in R n+1 ,  1 ≤ i ≤ n − 1,
                           i

                         C = C j + a jn C n+1 ,  1 ≤ j ≤ n − 1,
                           j
          when performed on B n+1 , result in the elements a ij and a ji being
                                 ∗
          transformed into a ∗  and a , where
                          ij     ji
                          ∗
                         a = a ij − a in + a jn ,  1 ≤ i ≤ n − 1,
                          ij
                         a = a ji − a jn + a in ,  1 ≤ j ≤ n − 1,
                          ∗
                          ji
                                 ∗
                            = −a .
                                 ij
          In particular, a ∗  = 0, so that all the elements except the last in both
                       in
          column n and row n are reduced to zero. Hence, when a Laplace expansion
          from the last two rows or columns is performed, only one term survives and
          the formula
                                            ∗
                                   B n+1 = |a | n−1
                                            ij
          emerges, which proves the theorem. When n is even, both sides of this
          formula are identically zero.


          4.3.2  Preparatory Lemmas

          Let



          where                      B n = |b ij | n
                                      1,    i<j − 1

                                b ij =  0,  i = j − 1
                                      −1, i>j − 1.
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