A permutation $\sigma$ is a bijection from the set $[n]=\{1,2,\dots ,n\}$ to itself. We denote it using the notation
$$\begin{pmatrix}1 & 2 & … & n \\ {\sigma(1)} & {\sigma(2)} &\dots & {\sigma(n)} \end{pmatrix},$$
e.g. if $n=3$ then $\begin{pmatrix}1&2&3 \\ 2&3&1 \end{pmatrix}$ denotes the permutation defined by $\sigma(1)=2, \sigma(2)=3$ and $\sigma(3)=1.$
An inversion in $\sigma$ is a pair $(i,j)$ such that $i<j$ but $\sigma(i)>\sigma(j).$ The sign of a permutation $\sigma$ (denoted $sgn(\sigma)$) is defined to be $(-1)^{inv(\sigma)}$, where $inv(\sigma)$ denotes the total number of inversions in $\sigma .$ In the above example, there are 2 inversions corresponding to the pairs $(1,3)$ and $(2,3)$ so that $sgn(\sigma)=(-1)^2-1.$ For each permutation $\sigma$, define a matrix $A_\sigma$ as follows:
$$A_{\sigma}(i,j) = \left\{ \begin{array}{rcl} 1 & \text{if } \sigma(i)=j \\ 0 & \text{otherwise } \end{array}\right.$$
Find $sgn(\sigma), sgn(\tau),A_{\sigma}$ and $A_{\tau}$ for the following permutations:
$$\sigma =\begin{pmatrix} 1 & 2 & 3&4&5&6 \\ 2& 3&4&5&1&6 \end{pmatrix}, \; \tau=\begin{pmatrix} 1 & 2 & 3&4&5&6 \\ 4& 1&3&2&6&5 \end{pmatrix}$$
