If a relation R is defined from a set `A={2, 3, 4, 5}` to a set `B={3, 6, 7, 10}` as follows `(x, y) in R hArr x` divides y. Expression of `R^(-1)` is represented by

Options

A. `{(6, 2), (10, 2), (3, 3)}`
B. `{(6, 2), (10, 5), (3, 3)}`
C. `{(6, 2), (10, 2), (3,3), (6, 3), (10, 5)}`
D. none of these

Answer

To find the inverse of the relation \( R \) defined from set \( A = \{2, 3, 4, 5\} \) to set \( B = \{3, 6, 7, 10\} \) where \( (x, y) \in R \) if \( x \) divides \( y \), we will follow these steps:

### Step 1: Identify the ordered pairs in relation \( R \)

We need to find all pairs \( (x, y) \) such that \( x \) divides \( y \) for \( x \in A \) and \( y \in B \).

- For \( x = 2 \):
- \( 2 \) divides \( 6 \) (since \( 6 \div 2 = 3 \))
- \( 2 \) divides \( 10 \) (since \( 10 \div 2 = 5 \))
- Ordered pairs: \( (2, 6) \), \( (2, 10) \)

- For \( x = 3 \):
- \( 3 \) divides \( 3 \) (since \( 3 \div 3 = 1 \))
- \( 3 \) divides \( 6 \) (since \( 6 \div 3 = 2 \))
- Ordered pairs: \( (3, 3) \), \( (3, 6) \)

- For \( x = 4 \):
- \( 4 \) does not divide \( 3 \), \( 6 \), \( 7 \), or \( 10 \).
- Ordered pairs: None

- For \( x = 5 \):
- \( 5 \) divides \( 10 \) (since \( 10 \div 5 = 2 \))
- Ordered pairs: \( (5, 10) \)

Combining all the ordered pairs, we have:
\[ R = \{(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)\} \]

### Step 2: Find the inverse relation \( R^{-1} \)

To find the inverse relation \( R^{-1} \), we switch the positions of \( x \) and \( y \) in each ordered pair of \( R \).

- From \( (2, 6) \) we get \( (6, 2) \)
- From \( (2, 10) \) we get \( (10, 2) \)
- From \( (3, 3) \) we get \( (3, 3) \)
- From \( (3, 6) \) we get \( (6, 3) \)
- From \( (5, 10) \) we get \( (10, 5) \)

Thus, the inverse relation \( R^{-1} \) is:
\[ R^{-1} = \{(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)\} \]

### Final Expression of \( R^{-1} \)

The expression of \( R^{-1} \) is:
\[ R^{-1} = \{(6, 2), (10, 2), (3, 3), (6, 3), (10, 5)\} \]