Statement-1: If R is an equivalence relation on a set A, then `R^(-1)` is also an equivalence relation.
Statement-2: `R = R^(-1)` iff R is a symmetric relation.

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for statement-1.
B. Statement-1 is True, Statement-2 is true, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given statements, we need to analyze each statement regarding equivalence relations and symmetric relations step by step.

### Step 1: Understanding Statement 1
**Statement 1:** If \( R \) is an equivalence relation on a set \( A \), then \( R^{-1} \) is also an equivalence relation.

**Solution:**
1. An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity.
2. Let \( R \) be an equivalence relation on set \( A \). This means:
- **Reflexive:** For every \( a \in A \), \( (a, a) \in R \).
- **Symmetric:** If \( (a, b) \in R \), then \( (b, a) \in R \).
- **Transitive:** If \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \).

3. Now, consider the inverse relation \( R^{-1} \):
- **Reflexive:** Since \( (a, a) \in R \), it follows that \( (a, a) \in R^{-1} \) for all \( a \in A \).
- **Symmetric:** If \( (a, b) \in R^{-1} \), then \( (b, a) \in R \) (which is true since \( R \) is symmetric).
- **Transitive:** If \( (a, b) \in R^{-1} \) and \( (b, c) \in R^{-1} \), then \( (b, a) \in R \) and \( (c, b) \in R \). By transitivity of \( R \), \( (c, a) \in R \), which means \( (a, c) \in R^{-1} \).

4. Since \( R^{-1} \) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation.

**Conclusion:** Statement 1 is true.

### Step 2: Understanding Statement 2
**Statement 2:** \( R = R^{-1} \) if and only if \( R \) is a symmetric relation.

**Solution:**
1. **(If Part)** Assume \( R \) is symmetric. By definition, if \( (a, b) \in R \), then \( (b, a) \in R \). This means that for every pair in \( R \), the corresponding pair in \( R^{-1} \) exists. Thus, \( R = R^{-1} \).

2. **(Only If Part)** Assume \( R = R^{-1} \). This means that for every \( (a, b) \in R \), it must also hold that \( (b, a) \in R \) because \( (a, b) \in R^{-1} \). Hence, \( R \) is symmetric.

**Conclusion:** Both directions of the statement hold true, so Statement 2 is also true.

### Final Conclusion
Both statements are true:
- Statement 1 is true: \( R^{-1} \) is an equivalence relation if \( R \) is an equivalence relation.
- Statement 2 is true: \( R = R^{-1} \) if and only if \( R \) is symmetric.