Let A=[ … ]. Then the number of 3× 3 matrices B with entries from the set 1,2,3,4,5 and satisfying AB=BA is .

Let, the matrix $B=[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}]$B=[adg​beh​cfi​]

$$\because AB=BA$$∵AB=BA

$$[\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{matrix}][\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}]=[\begin{matrix}a & b & c \\ d & e & f \\ g & h & i\end{matrix}][\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{matrix}]$$[010​100​001​][adg​beh​cfi​]=[adg​beh​cfi​][010​100​001​]

$$\Rightarrow [\begin{matrix}d & e & f \\ a & b & c \\ g & h & i\end{matrix}]=[\begin{matrix}b & a & c \\ e & d & f \\ h & g & i\end{matrix}]$$⇒[dag​ebh​fci​]=[beh​adg​cfi​]

On equating the matrices, we get

$$\Rightarrow d=b,e=a,f=c,g=h$$⇒d=b,e=a,f=c,g=h

$\therefore$∴ Matrix $B=[\begin{matrix}a & b & c \\ b & a & c \\ g & g & i\end{matrix}]$B=[abg​bag​cci​]

No. of ways of selecting each of $a,b,c,g,i$a,b,c,g,i from the set ${1,2,3,4,5}$1,2,3,4,5 are $5.$5.

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

$$={5}^{5}=3125$$=55=3125

$\therefore$∴ No. of Matrices $B=3125.$B=3125.