Let a relation R be defined by relation The `R = {(4, 5), (1,4), (4, 6), (7, 6), (3,7)}` `R^(-1)oR` is given by

Answer

To solve the problem, we need to find \( R^{-1} \) (the inverse of the relation \( R \)) and then compute \( R^{-1} \circ R \) (the composition of \( R^{-1} \) and \( R \)).

### Step 1: Find the inverse relation \( R^{-1} \)

The inverse of a relation \( R \) is obtained by swapping the elements in each ordered pair.

Given:
\[ R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\} \]

The inverse relation \( R^{-1} \) will be:
\[ R^{-1} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\} \]

### Step 2: Compute the composition \( R^{-1} \circ R \)

The composition of two relations \( A \) and \( B \), denoted \( A \circ B \), consists of all pairs \( (a, c) \) such that there exists an element \( b \) where \( (a, b) \in B \) and \( (b, c) \in A \).

Now we will find \( R^{-1} \circ R \):
1. For each pair \( (x, y) \) in \( R \), we will look for pairs \( (y, z) \) in \( R^{-1} \).
2. If such a pair exists, we will form the new pair \( (x, z) \).

Let's go through each pair in \( R \):

- For \( (4, 5) \) in \( R \):
- Look for \( (5, z) \) in \( R^{-1} \): \( (5, 4) \) gives \( (4, 4) \).

- For \( (1, 4) \) in \( R \):
- Look for \( (4, z) \) in \( R^{-1} \): \( (4, 1) \) gives \( (1, 1) \).

- For \( (4, 6) \) in \( R \):
- Look for \( (6, z) \) in \( R^{-1} \): \( (6, 4) \) gives \( (4, 4) \) and \( (6, 7) \) gives \( (4, 7) \).

- For \( (7, 6) \) in \( R \):
- Look for \( (6, z) \) in \( R^{-1} \): \( (6, 4) \) gives \( (7, 4) \) and \( (6, 7) \) gives \( (7, 7) \).

- For \( (3, 7) \) in \( R \):
- Look for \( (7, z) \) in \( R^{-1} \): \( (7, 3) \) gives \( (3, 3) \).

Now, compiling all the pairs we found:
- From \( (4, 5) \): \( (4, 4) \)
- From \( (1, 4) \): \( (1, 1) \)
- From \( (4, 6) \): \( (4, 4) \) and \( (4, 7) \)
- From \( (7, 6) \): \( (7, 4) \) and \( (7, 7) \)
- From \( (3, 7) \): \( (3, 3) \)

Thus, the final result for \( R^{-1} \circ R \) is:
\[ R^{-1} \circ R = \{(4, 4), (1, 1), (4, 7), (7, 4), (7, 7), (3, 3)\} \]

### Final Answer:
\[ R^{-1} \circ R = \{(4, 4), (1, 1), (4, 7), (7, 4), (7, 7), (3, 3)\} \]