A permutation matrix is a square matrix obtained from the same size identity matrix by a
permutation of rows. Such a matrix is always row equivalent to an identity.
Every row and every column of a permutation matrix contain exactly one nonzero entry,
which is $1.$
Every permutation matrix is a product of elementary row-interchange matrices. The elementary matrix factors may be chosen to only involve adjacent rows.
- A general permutation matrix is not symmetric.
- A general permutation matrix does not agree with its inverse.
- A product of permutation matrices is again a permutation matrix.
An $n \times n$ permutation matrix is a matrix obtained from the $n \times n$ identity matrix by permuting its rows.
Given that $P = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}$
If $\sigma$ is the permutation the matrix encodes, then $\sigma(i)$ is given by the column index of the entry containing the $1$ in row $i.$ For above $\sigma$ the permutation would be $\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3\end{pmatrix}$ because in row $1,$ column $2$ contains $1;$ in row $2,$ column $4$ contains $1;$ in row $3,$ column $1$ contains $1;$ in row $4,$ column $3$ contains $1.$
Given an $n \times n$ permutation matrix $P$ encoding the permutation $\sigma,$ the determinant of $P$ is simply $(-1)^{\#\text{inv}(\sigma)}.$ For the example above, there are three inversions. So the determinant is $(-1)^{3} = -1.$
This can be readily seen from the definition of the determinant: As each term in the definition consists of $(-1)^{\#\text{inv}(\sigma')}$ for some permutation $\sigma'$ times the product of $n$ entries from the matrix, exactly one from each row and one from each column, the only way we get a nonzero term from $P$ is to have a permutation that picks the $1$ from each row. The only permutation that does that is $\sigma.$
So, the correct answer is $-1.$
