# 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.