Let $$A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$. Then the number of $$3 \times 3$$ matrices $$B$$ with entries from the set $$\{1, 2, 3, 4, 5\}$$ and satisfying $$AB = BA$$ is ___.

The matrix $$A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is a permutation matrix that swaps rows 1 and 2 when it left-multiplies a matrix, and swaps columns 1 and 2 when it right-multiplies a matrix.

Let $$B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$. Then $$AB$$ is obtained by swapping rows 1 and 2 of $$B$$: $$AB = \begin{bmatrix} b_{21} & b_{22} & b_{23} \\ b_{11} & b_{12} & b_{13} \\ b_{31} & b_{32} & b_{33} \end{bmatrix}$$.

And $$BA$$ is obtained by swapping columns 1 and 2 of $$B$$: $$BA = \begin{bmatrix} b_{12} & b_{11} & b_{13} \\ b_{22} & b_{21} & b_{23} \\ b_{32} & b_{31} & b_{33} \end{bmatrix}$$.

Setting $$AB = BA$$, we compare entry by entry:

From position $$(1,1)$$: $$b_{21} = b_{12}$$. From $$(1,2)$$: $$b_{22} = b_{11}$$. From $$(1,3)$$: $$b_{23} = b_{13}$$.

From $$(2,1)$$: $$b_{11} = b_{22}$$ (same as above). From $$(2,2)$$: $$b_{12} = b_{21}$$ (same as above). From $$(2,3)$$: $$b_{13} = b_{23}$$ (same as above).

From $$(3,1)$$: $$b_{31} = b_{32}$$. From $$(3,2)$$: $$b_{32} = b_{31}$$ (same). From $$(3,3)$$: $$b_{33} = b_{33}$$ (always true).

So the independent constraints are: $$b_{11} = b_{22}$$, $$b_{12} = b_{21}$$, $$b_{13} = b_{23}$$, and $$b_{31} = b_{32}$$.

The free parameters are: $$b_{11}$$ (which determines $$b_{22}$$), $$b_{12}$$ (which determines $$b_{21}$$), $$b_{13}$$ (which determines $$b_{23}$$), $$b_{31}$$ (which determines $$b_{32}$$), and $$b_{33}$$. That gives 5 free parameters, each taking values from $$\{1, 2, 3, 4, 5\}$$.

The total number of such matrices is $$5^5 = 3125$$.