Inverse of Permutations

If f be a permutation of degree n, defined on a finite set S consisting of n distinct elements, by definition f is a one-one mapping of S onto itself. Since f is one-one onto, it is invertible. Let {f^{ - 1}} be the inverse of map f then {f^{ - 1}} will also be one-one map of S onto itself. Thus,{f^{ - 1}} is also a permutation of degree n on S. This {f^{ - 1}} is known as the inverse of the permutation f.

Thus if

 f = \left( {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}& \ldots &{{a_n}} \\ {{b_1}}&{{b_2}}&{{b_3}}& \ldots &{{b_n}} \end{array}} \right)


 {f^{ - 1}} = \left( {\begin{array}{*{20}{c}}{{b_1}}&{{b_2}}&{{b_3}}& \ldots &{{b_n}} \\ {{a_1}}&{{a_2}}&{{a_3}}& \ldots &{{a_n}} \end{array}} \right)

Note: Evidently {f^{ - 1}} is obtained by interchanging the rows of f because f\left( {{a_1}} \right) = {b_1} \Rightarrow {f^{ - 1}}\left( {{b_1}} \right) = {a_1} etc.