Let A = [ cases 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 cases ] . Then the...

Let, the matrix B = \left[\begin{cases}a & b & c \\ d & e & f \\ g & h & i\end{cases}\right]

$∵ AB = BA$

\left[$\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}$\right]\left[$\begin{cases}a & b & c \\ d & e & f \\ g & h & i\end{cases}$\right] = \left[$\begin{cases}a & b & c \\ d & e & f \\ g & h & i\end{cases}$\right]\left[$\begin{cases}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>$⇒ \left[$\begin{cases}d & e & f \\ a & b & c \\ g & h & i\end{cases}$\right] = \left[$\begin{cases}b & a & c \\ e & d & f \\ h & g & i\end{cases}$\right]$</p><p>On equating the matrices, we get</p><p>$⇒ d = b, e = a, f = c, g = h$</p><p>$∴$Matrix$B = \left[$\begin{cases}a & b & c \\ b & a & c \\ g & g & i\end{cases}$\right]$</p><p>No. of ways of selecting each of$a, b, c, g, i$from the set \left\{1, 2, 3, 4, 5\right\}$ are $5 .$

Thus, the total number of ways are $= 5 \times 5 \times 5 \times 5 \times 5$

$= {5}^{5} = 3125$

$∴$ No. of Matrices $B = 3125 .$