Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then the number of $3 \times 3$…
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 ________.
Solution
$B = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$
$AB = BA$
$\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} d & e & f \\ a & b & c \\ g & h & i \end{bmatrix} = \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}$
On equating the matrices, we get
$\Rightarrow d = b, \quad e = a, \quad f = c, \quad g = h$
$\therefore$ Matrix $B = \begin{bmatrix} a & b & c \\ b & a & c \\ g & g & i \end{bmatrix}$
No. of ways of selecting each of $a, b, c, g, i$ from the set $\{1, 2, 3, 4, 5\}$ are $5$.
Thus, the total number of ways are $= 5 \times 5 \times 5 \times 5 \times 5$
$= 5^5 = 3125$
$\therefore$ No. of Matrices $B = 3125$.