# 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

Then

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.