Animated Solution for Mathematics - Matrices and Determinants: Let $A=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$. Then the number of $3\times 3$ matrices $B$ with entries from the set $\{1,2,3,4,5\}$ and satisfying $AB=BA$ is

Understanding Commutation

Welcome, future engineer. Today, we are not just solving a matrix problem; we are uncovering the hidden geometry of linear algebra.

When you see a matrix like do not just see a grid of numbers. See an operator. See a transformation. This matrix is a permutation matrix, and it holds the power to rearrange the very structure of any matrix it touches.

The Operator Perspective

Imagine you have a general matrix We want to find how many such matrices $B$ exist, given that their entries are chosen from the set $\{1,2,3,4,5\}$, such that $AB=BA$.

If you try to multiply these matrices blindly, you will drown in a sea of variables. Instead, let us look at what $A$ actually does.

When you perform $AB$, you are multiplying $B$ by $A$ on the left. Because $A$ is essentially an identity matrix with the first two rows swapped, the product $AB$ will simply swap the first and second rows of $B$:

Now, consider $BA$. When you multiply by $A$ on the right, you are performing a column operation. Specifically, it swaps the first and second columns of $B$:

The Constraint Hunt

Now, we invoke the condition $AB=BA$. For these two matrices to be equal, every single corresponding entry must match.

Comparing the first rows, we get $d=b$, $e=a$, and $f=c$. These are our first three constraints.

If we look at the second rows, we find $a=e$, $b=d$, and $c=f$. Notice that these are identical to the constraints we already found; the system is consistent, and we have not lost any information.

Finally, we look at the third row. We have $g=h$, $h=g$, and $i=i$. The equation $i=i$ is a tautology—it is always true, regardless of what value $i$ takes. The only new constraint here is $g=h$.

The Combinatorial Finale

We have successfully reduced the complexity of our matrix $B$. By substituting our constraints, the matrix $B$ now looks like this:

Look at the variables remaining. We have $a,b,c,g,$ and $i$. These are our five independent degrees of freedom. Every other entry in the matrix is locked into place by these five.

Since each of these five variables can be chosen from the set $\{1,2,3,4,5\}$, we have 5 choices for $a$, 5 for $b$, 5 for $c$, 5 for $g$, and 5 for $i$. The total number of such matrices is simply the product of these choices:

There you have it. We did not need brute force. We needed insight. We needed to see the matrix not as a static object, but as a dynamic operator. Keep this perspective, and the final answer is 3125.