How to find a $3\times 3$ permutation matrix? Could anyone tell me how to find a $3\times 3$ permutation matrix $P$ such that for any $3\times n$ matrix $A$ the matrix $PA$ is $A$ with its last two rows exchanged.
I am simply not understanding the problem at all. thanks for helping.
 A: Use $P= \begin{bmatrix}
1 & 0 & 0  \\
0 & 0 &1  \\
0 & 1 & 0  \\
\end{bmatrix}.
$ 
Then for $A=
\begin{bmatrix}
a & b & c  \\
d & e &f  \\
g & h & i  \\
\end{bmatrix}
$  we have:
$$PA=\begin{bmatrix}
1 & 0 & 0  \\
0 & 0 &1  \\
0 & 1 & 0  \\
\end{bmatrix}\cdot \begin{bmatrix}
a & b & c  \\
d & e &f  \\
g & h & i  \\
\end{bmatrix} = \begin{bmatrix}
a & b & c  \\
g& h &i  \\
d & e & f  \\
\end{bmatrix} =A_{\text{swapped}} $$
The first column of $P$ ensures that the first row of $A$ does not change. The other two columns then swap the second and third row. If you swapped the second and third column in $P$ then $P$ would be the identity matrix and all rows in $A$ (and the columns too) would remain in their place.
For arbitrary $A\in \mathbb{R}^{3\times n}$ the result is the same.
A: Left-multiplying a matrix by the $i$th row of the appropriately-sized identity matrix picks out its $i$th row. Similarly, right-multiplying by the $j$th column of the identity picks out the $j$th column of the matrix.  
You want to leave the first row of your matrix alone, so the first row of the permutation matrix is $\small{\begin{bmatrix}1&0&0\end{bmatrix}}$. You’re swapping the second and third rows, i.e., the second row of the result should be the third row of the original matrix, so the second row of the permutation matrix is $\small{\begin{bmatrix}0&0&1\end{bmatrix}}$. Can you finish this on your own?
