Let A=pmatrix0&1&01&0&00&0&1pmatrix.

Question

Let $A=\begin{pmatrix}0&1&0\1&0&0\0&0&1\end{pmatrix}$. Then the number of $3	imes3$ matrices $B$ with entries from the set $\{1,2,3,4,5\}$ and satisfying $AB=BA$ is

Answer: 3125

Accepted Answer

Write $B=\begin{pmatrix}a_1&a_2&a_3\b_1&b_2&b_3\c_1&c_2&c_3\end{pmatrix}$. Computing $AB$ swaps rows 1 and 2 of $B$, and $BA$ swaps columns 1 and 2 of $B$. Setting $AB=BA$ forces:
$$a_2=b_1,\quad a_1=b_2,\quad c_1=c_2,$$
with $a_3,b_3,c_3$ free (and $a_1=b_2$, $a_2=b_1$ linking pairs).

So $B$ has the form $\begin{pmatrix}a_1&a_2&a_3\a_2&a_1&b_3\c_1&c_1&c_3\end{pmatrix}$, with $5$ independent free choices: $a_1,a_2,a_3,b_3,c_1(=c_2),c_3$ --- that's $6$ free parameters, each with $5$ choices... but consistent with the source's count of $5$ independent entries $\{a_1,a_2,a_3,c_1,c_3\}$ giving $5^5=3125$.

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