Page 83 - Determinants and Their Applications in Mathematical Physics
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68   4. Particular Determinants

          Proof. Let A n = |a ij | n and let E n+1 and F n+1 denote determinants
          obtained by bordering A n in different ways:
                                 1    1     1     1   ···

                                 −x   •     a 12  a 13  ···


                        E n+1 =   −x −a 12  •    a 23  ···

                                 −x −a 13  −a 23  •   ···

                                ···  ···    ···  ···  ···
                                                         n+1
          and F n+1 is obtained by replacing the first column of E n+1 by the column

                                                 T
                                0 − 1 − 1 − 1 ···   .
                                                 n+1
          Both A n and F n+1 are skew-symmetric. Then,
                                 E n+1 = A n + xF n+1 .
          Return to E n+1 and perform the column operations
                            C = C j − C 1 ,  2 ≤ j ≤ n +1,

                              j
          which reduces every element to zero except the first in the first row and
          increases every other element in columns 2 to (n +1) by x. The result is
                                  E n+1 = |a ij + x| n .
          Hence, applying Theorems 4.9 and 4.10,

                               |a ij + x| 2n = A 2n + xF 2n+1
                                         = A 2n ,

                             |a ij + x| 2n−1 = A 2n−1 + xF 2n
                                         = xF 2n .
          The theorem follows.
          Corollary. The determinant
                          A = |a ij | 2n ,  where  a ij + a ji =2x,

          can be expressed as a skew-symmetric determinant of the same order.
          Proof. The proof begins by expressing A in the form
                             x

                                     a 12     a 13   a 14  ···
                                      x
                           2x − a 12          a 23   a 24  ···

                    A =   2x − a 13  2x − a 23  x    a 34  ···

                           2x − a 14  2x − a 24  2x − a 34  x  ···

                            ···      ···       ···    ···  ···
                                                              2n
          and is completed by subtracting x from each element.
            Let
                               A n = |a ij | n ,  a ji = −a ij ,
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