Consider the matrix$$ P=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$ that corresponds to permuting the columns of the identity matrix under the permutation $(1 2 4 3)$ (i.e., the first column of the identity matrix is the second column of $P$ , the second column of the identity matrix is the fourth column of $P$ , etc.). Then $P$ is a permutation matrix.

We will consider what happens when we multiply a $4\times 4$ matrix by $P$ . For example, let $A$ be the matrix$$ A=\begin{pmatrix} 4 & 2 & 6 & 8 \\ 1 & 3 & 5 & 7 \\ 1 & 0 & 1 & 0 \\ -1& -2& -3& -4 \end{pmatrix}.$$ Then$$ PA=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 4 & 2 & 6 & 8 \\ 1 & 3 & 5 & 7 \\ 1 & 0 & 1 & 0 \\ -1& -2& -3&-4 \end{pmatrix} = \begin{pmatrix} 1 & 3 & 5 & 7 \\ -1& -2& -3&-4 \\ 4 & 2 & 6 & 8 \\ 1 & 0 & 1 & 0 \end{pmatrix}.$$

We notice that $PA$ has the same rows as $A$ . Moreover, the rows of $A$ are the rows of $PA$ permuted under $(1 2 4 3)$ : The first row of $PA$ is the second row of $A$ , the second row of $PA$ is the fourth row of $A$ , etc.