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Find determinant of the $n \times n$ permutation matrix $$ M= \left[ {\begin{array}{cccc} 0 & 0 & \ldots & 0 & 1\\ 0 & 0 & \ldots & 1 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 1 & 0 & \ldots & 0 & 0\\ \end{array} } \right] $$

My answer was $(-1)^n$. Am I correct

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  • $\begingroup$ ohh so it is always $-1$ $\endgroup$
    – Bhaskar Vashishth
    Dec 21, 2014 at 18:58
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    $\begingroup$ Consider the difference between "even" and "odd" permutations, and then consider how many row transpositions it would take to convert an even/odd permutation matrix to the identity. $\endgroup$
    – apnorton
    Dec 21, 2014 at 19:01
  • $\begingroup$ Hmmm... $(-1)^{\lfloor n/2 \rfloor}$ $\endgroup$
    – Bhaskar Vashishth
    Dec 21, 2014 at 19:01

2 Answers 2

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Remember cofactor expansion: $$\det \left[ {\begin{matrix} {\color{red}0} & {\color{red}0} & \ldots & {\color{red}0} & {\color{red}1}\\ 0 & 0 & \ldots & 1 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 1 & \ldots & 0 & 0\\ 1 & 0 & \ldots & 0 & 0\\ \end{matrix} } \right] = $$

$${\color{red}0}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 1\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] - {\color{red}0}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 1\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] + {\color{red}0}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 1\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] +\cdots $$

$$\cdots + (-1)^{n-2}{\color{red}0}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 0\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] + (-1)^{n-1}{\color{red}1}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 1\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] $$

So the only thing that survives is

$$(-1)^{n-1}\cdot \det \left[ {\begin{matrix} 0 & \ldots & 0 & 1\\ \vdots & \ddots & \vdots & \vdots\\ 0 & \ldots & 0 & 0\\ 1 & \ldots & 0 & 0\\ \end{matrix} } \right] $$

Performing this $n$ times you will get $(-1)^{n(n-1)/2}$

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  • $\begingroup$ The option were $1,-1, (-1)^n, (-1)^{\lfloor n/2 \rfloor}$ $\endgroup$
    – Bhaskar Vashishth
    Dec 21, 2014 at 19:36
  • $\begingroup$ If $n=2k$ is even for example, $\lfloor n/2\rfloor = k$ and $n(n-1)/2 = k(2k-1)$. If $k$ is even then $k(2k-1)$ is even. If $k$ is odd then so is $k(2k-1)$. So minus 1 raised to either gives the same result. The same is true if $n$ is odd, which Im sure you can do now. $\endgroup$
    – David P
    Dec 21, 2014 at 19:41
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i think the answer is $det\ \pmatrix{0 & 0 & \cdots & 0 & 1\\0 & 0& \cdots & 1 & 0\\ \vdots& \vdots& \ddots&\vdots&\vdots\\1&0&\cdots&0&0} = \left\{ \begin{array}{ll} (-1)^{(n-1)/2} & \mbox{ if $n$ is odd }\\ (-1)^{n/2} & \mbox{ if $n$ is even} \end{array} \right.$

here is how you see this. let $\lambda$ be an eigenvalue of $M.$ then you have $$\{x_n = \lambda x_1, x_1 = \lambda x_n\}, \{x_{n-1} = \lambda x_2, x_2 = \lambda x_{n-1}\}, \cdots $$ these come in pairs except when $n$ is odd you are left with single equation for the middle $x_{(n+1)/2}$. for each pair you get the the eigenvalues $\pm 1$ and $1$ for the single one. now the determinant of $M$ is the product of $-1$ for each pair, hence the formula.

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