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5.6 Distinct Matrices with Nondistinct Determinants  217

                                           1    −x     x  2
                    φ 1 (x + y) φ 2 (x + y)


                                       = φ 0 (y) φ 1 (y) φ 2 (y) ,
                    φ 2 (x + y) φ 3 (x + y)
                                           φ 1 (y) φ 2 (y) φ 3 (y)

                                           1     −y     y  2

                                      = φ 0 (x) φ 1 (x) φ 2 (x)     (5.6.19)


                                           φ 1 (x) φ 2 (x) φ 3 (x)
                                               1     −x     x
                                                             2


                    φ 2 (x + y) φ 3 (x + y)       1  φ 0 (y) φ 1 (y) φ 2 (y)
                                       =
                    φ 3 (x + y) φ 4 (x + y)       −xφ 1 (y) φ 2 (y) φ 3 (y)
                                         x   φ 2 (y) φ 3 (y) φ 4 (y)
                                           2
                                                 1     −2x    3x
                                                                2

                                           1    −x     x
                                                         2      3
                                                                    .(5.6.20)
                                                             −x
                                          φ 0 (y) φ 1 (y) φ 2 (y)  φ 3 (y)
                                      =
                                         φ 1 (y) φ 2 (y) φ 3 (y)  φ 4 (y)

          Do these identities possess duals?
          5.6.3  Determinants with Stirling Elements
          Matrices s n (x) and S n (x) whose elements contain Stirling numbers of
          the first and second kinds, s ij and S ij , respectively, are defined in
          Appendix A.1.
            Let the matrix obtained by rotating S n (x) through 90 ◦  in the

          anticlockwise direction be denotes by S n (x). For example,
                                                    1
                                                      
                                               1
                                                  10x 
                                          1   6x   25x  .
                                                     2 
                           S 5 (x)= 
                                      1  3x 7x     15x
                                               2     3 
                                   1  x  x 2  x 3   x 4
          Define another nth-order triangular matrix B n (x) as follows:
                                  ←
                        B n (x)=[b ij x i−j ],  n ≥ 2,  1 ≤ i, j ≤ n,
          where
                           j−1
                      1              j − 1           i−1
              b ij =         (−1) r        (n − r − 1)  ,  i ≥ j.   (5.6.21)
                   (j − 1)!           r
                           r=0
          These numbers are integers and satisfy the recurrence relation
                             b ij = b i−1,j−1 +(n − j)b i−j,j ,
          where
                                       b 11 =1.                     (5.6.22)
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