Let S be a set of all square matrices of order 2 . If a relation R defined on set S such that `ARB=>AB=BA`, then relation R is `(A,BepsilonS)`

Explanation

To determine the properties of the relation \( R \) defined on the set \( S \) of all square matrices of order 2, where \( A R B \) if and only if \( AB = BA \), we will check if \( R \) is reflexive, symmetric, and transitive.

### Step 1: Check for Reflexivity
A relation \( R \) is reflexive if for every element \( A \in S \), it holds that \( A R A \).

- For any matrix \( A \), we have:
\[ A R A \implies AA = AA \]
- This statement is always true since the product of a matrix with itself is equal.

**Conclusion**: The relation \( R \) is reflexive.

### Step 2: Check for Symmetry
A relation \( R \) is symmetric if for every \( A, B \in S \), if \( A R B \) then \( B R A \).

- Assume \( A R B \) holds, which means:
\[ AB = BA \]
- We need to show that \( B R A \) holds, i.e., \( BA = AB \).
- Since \( AB = BA \) is given, it follows that \( BA = AB \) is also true.

**Conclusion**: The relation \( R \) is symmetric.

### Step 3: Check for Transitivity
A relation \( R \) is transitive if for every \( A, B, C \in S \), if \( A R B \) and \( B R C \) then \( A R C \).

- Assume \( A R B \) and \( B R C \):
\[ AB = BA \quad \text{(1)} \]
\[ BC = CB \quad \text{(2)} \]
- We need to check if \( A R C \) holds, i.e., if \( AC = CA \).
- From (1) and (2), we cannot directly conclude that \( AC = CA \) holds for all matrices \( A, B, C \). There are cases where \( A \) and \( C \) do not commute even if \( A \) commutes with \( B \) and \( B \) commutes with \( C \).

**Conclusion**: The relation \( R \) is not transitive.

### Final Conclusion
The relation \( R \) is reflexive and symmetric but not transitive.

---