If `R sub A xx B` and `S sub B xx C` be two relations, then `(SoR)^-1 =`

Options

A. `S^(-1)oR^(-1)`
B. `R^(-1)oS^(-1)`
C. SoR
D. RoS

Answer

To solve the problem, we need to find the inverse of the composition of two relations \( S \) and \( R \). Given that \( R \subseteq A \times B \) and \( S \subseteq B \times C \), we want to find \( (S \circ R)^{-1} \).

### Step-by-Step Solution:

1. **Understanding Relations**:
- We have two relations: \( R \) from set \( A \) to set \( B \) and \( S \) from set \( B \) to set \( C \).
- The composition \( S \circ R \) is defined as \( S \circ R = \{ (a, c) \mid \exists b \in B \text{ such that } (a, b) \in R \text{ and } (b, c) \in S \} \).

**Hint**: Remember that the composition of relations involves finding a common element in the sets of the relations.

2. **Finding the Inverse of the Composition**:
- The inverse of a relation \( T \) is defined as \( T^{-1} = \{ (y, x) \mid (x, y) \in T \} \).
- Therefore, the inverse of the composition \( (S \circ R)^{-1} \) can be expressed as:
\[ (S \circ R)^{-1} = \{ (c, a) \mid (a, c) \in S \circ R \} \]

**Hint**: Recall how to define the inverse of a relation.

3. **Using the Definition of Composition**:
- From the definition of composition, if \( (a, c) \in S \circ R \), then there exists some \( b \) such that \( (a, b) \in R \) and \( (b, c) \in S \).
- Thus, we can write:
\[ (c, a) \in (S \circ R)^{-1} \text{ if and only if } (a, b) \in R \text{ and } (b, c) \in S \]

**Hint**: Think about how the elements relate to each other in terms of their positions in the relations.

4. **Expressing the Inverses**:
- From the above, we can see that:
- If \( (b, c) \in S \), then \( (c, b) \in S^{-1} \).
- If \( (a, b) \in R \), then \( (b, a) \in R^{-1} \).
- Therefore, we can express:
\[ (c, a) \in (S \circ R)^{-1} \text{ if and only if } (c, b) \in S^{-1} \text{ and } (b, a) \in R^{-1} \]

**Hint**: Use the property of inverses to relate elements in the original relations to their inverses.

5. **Final Expression**:
- This means that:
\[ (S \circ R)^{-1} = R^{-1} \circ S^{-1} \]
- Hence, we conclude that:
\[ (S \circ R)^{-1} = R^{-1} \circ S^{-1} \]

### Conclusion:
Thus, the final answer is:
\[ (S \circ R)^{-1} = R^{-1} \circ S^{-1} \]