I want to compute the inverse of:
$\begin{pmatrix} 1&2&3&4&5&6&7&8&9\\3&2&1&6&5&9&4&8&7 \end{pmatrix}$
Sorry about alignment(they are all one digit though)
Now it would seem that the inverse here is itself. But I feel like maybe I am just confusing myself now, and I want some confirmation. Thank you and sorry again.
Also can someone explain why my matrix above isn't formatting properly despite having two & between each element
On
Each element in $S_9$ is a permutation (or bijection) $\{1, 2, \dots, 9\} \to \{1, 2, \dots, 9\}$. In particular, your permutation I can denote as a function $f$, where:
$$\begin{align} f(1) &= 3\\ f(2) &= 2 \\ f(3) &= 1 \\ f(4) &= 6 \\ f(5) &= 5 \\ f(6) &= 9 \\ f(7) &= 4 \\ f(8) &= 8 \\ f(9) &= 7\text{.} \end{align}$$ To compute the inverse of your permutation, notice that the inverse of $f$ exists because it is a bijection, and thus $$\begin{align} f^{-1}(3) &= 1\\ f^{-1}(2) &= 2 \\ f^{-1}(1) &= 3 \\ f^{-1}(6) &= 4 \\ f^{-1}(5) &= 5 \\ f^{-1}(9) &= 6 \\ f^{-1}(4) &= 7 \\ f^{-1}(8) &= 8 \\ f^{-1}(7) &= 9\text{.} \end{align}$$ Hence $$\begin{pmatrix} 1&2&3&4&5&6&7&8&9\\3&2&1&6&5&9&4&8&7 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5& 6 & 7 & 8 & 9 \\ 3 & 2 & 1 & 7 & 5 & 4 & 9 & 8 & 6\end{pmatrix}\text{.}$$
To compute the inverse of a permutation that is written in two-line notation, simply swap the two rows, then reorder the columns so that the first row goes $1,2,3\cdots,n$.
To compute the inverse of a permutation that is written in disjoint cycle notation, simply write each cycle backwards. In fact, even if we have a representation as a product of non-disjoint cycles, simply write the whole thing in reverse (so $(12)(341)(2)(3)(4)$ becomes $(4)(3)(2)(143)(21)$ for example).