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Step-by-Step Solution
1. Understand the Problem
We are given the matrix
A = \begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{bmatrix},
and we need to find the number of all possible 3×3 matrices
B = \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}
with entries from the set {1, 2, 3, 4, 5} such that AB = BA .
2. Write the Condition AB = BA
Compute AB and BA :
AB = \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} d & e & f \\ a & b & c \\ g & h & i \end{bmatrix}.
BA = \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}
= \begin{bmatrix} b & a & c \\ e & d & f \\ h & g & i \end{bmatrix}.
Since AB = BA, we compare both products:
\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}.
3. Equate Corresponding Entries
From the equality of matrices, we find:
d = b
e = a
f = c
g = h
Therefore, the matrix B must be of the form:
B = \begin{bmatrix}
a & b & c \\
b & a & c \\
g & g & i
\end{bmatrix}.
4. Count the Number of Possible Matrices
Each of a, b, c, g, and i can be chosen independently from the set {1, 2, 3, 4, 5}. Thus, for each of these five positions, there are 5 possible choices. Hence, the total number of such matrices is:
5 \times 5 \times 5 \times 5 \times 5 = 5^5 = 3125.
5. Final Answer
The number of 3×3 matrices B satisfying AB = BA is \boxed{3125} .
