Answer
Corresponding permutation matrix $P$ to permutation $(2,3,...,1)$ is given by:
$P=\left[ {\begin{array}{cccc}
0&1&0&...&0\\
0&0&1&...&0\\
.&.&.&&.\\
.&.&.&&.\\
0&0&0&...&1\\
1&0&0&...&0\\
\end{array} } \right]$
$\text{since n-1 exchanges are required to restore (1,2,.......,n), it follows that if n is even)}$
then, $P$ is odd and $det{P}=-1$, while if $n$ is odd then $P$ is even and $det{P}=1.$
Work Step by Step
Corresponding permutation matrix $P$ to permutation $(2,3,...,1)$ is given by:
$P=\left[ {\begin{array}{cccc}
0&1&0&...&0\\
0&0&1&...&0\\
.&.&.&&.\\
.&.&.&&.\\
0&0&0&...&1\\
1&0&0&...&0\\
\end{array} } \right]$
$\text{since n-1 exchanges are required to restore (1,2,.......,n), it follows that if n is even)}$
This is because none of the numbers between these two permutations are in the same position and each exchange puts exactly one of them in its starting position .After $n-1 $exchanges the $nth$ will automatically fall in its starting position.$\textbf{Therefore, if $n$ is the even given permutation will be odd, while if $n$ is odd the permutation will be even. In the other words if$n$ is even then $detP=-1$ and if $n$ is odd then $detP=-1$}$
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