Page 217 - Determinants and Their Applications in Mathematical Physics
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202   5. Further Determinant Theory

                                   a 1 a 2 a 3 + a 1 b 3 + a 3 b 2
                         =                                  .
                            a 1 a 2 a 3 + a 1 b 3 + a 3 b 2 + a 2 a 3 b 1 + b 1 b 3
          Each of these fractions can be expressed in the form H 11 /H, where H is a
          tridiagonal determinant:

                                      |a 1 |
                                             ,
                              f 1 =
                                      1

                                         b 1
                                    −1 a 1


                                       a 1

                                           b 2
                                      −1 a 2

                                                 ,
                              f 2 =
                                      1
                                         b 1

                                    −1
                                         a 1
                                             b 2
                                        −1 a 2


                                       a 1  b 2


                                        −1  a 2
                                               b 3
                                           −1 a 3

                                                      .
                                     1   b 1
                              f 3 =
                                     −1  a 1  b 2

                                        −1

                                             a 2  b 3

                                             −1 a 3
          Theorem 5.9.
                                           (n+1)
                                         H 11
                                    f n =      ,
                                          H n+1
          where
                         1   b 1

                         −1  a 1  b 2

                             −1

                                  a 2  b 3
                                  .   .     .
                H n+1 =           . .  . .  . .               .      (5.5.2)

                                      −1

                                           a n−2  b n−1
                                           −1    a n−1

                                                       b n
                                                  −1   a n n+1

          Proof. Use the method of induction. Assume that
                                             (n)
                                           H 11
                                    f n−1 =  H n  ,
          which is known to be true for small values of n. Hence, adding b n /a n to
          a n−1 ,
                                            (n)
                                          K 11
                                     f n =  K n  ,                   (5.5.3)
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