# Inverse of Permutations

If $$f$$ is 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 a one-one mapping 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

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

then

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

**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.