Login/Sign Up
Home » Find the inverse relation R-1 in each of the following cases:
Find the inverse relation R-1 in each of the following cases:
CBSE | Class 11 | Exercise 2.3 | Maths | RD Sharma | Relations
Find the inverse relation R-1 in each of the following cases:

(i) R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R= {(x, y) : x, y ∈ N; x + 2y = 8}

Solution:

(i) It is given that:

R=\left\{ \left( 1,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 3,\text{ }2 \right),\text{ }\left( 5,\text{ }6 \right) \right\}

So, R‑1 is given as follows:

{{R}^{-1}}=\left\{ \left( 2,\text{ }1 \right),\text{ }\left( 3,\text{ }1 \right),\text{ }\left( 3,\text{ }2 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 6,\text{ }5 \right) \right\}

(ii) It is given that : R= {(x, y): x, y ∈ N; x + 2y = 8}

Here, we have: x + 2y = 8

Or, x = 8 – 2y

As y ∈ N, Put the values of y = 1, 2, 3,…… till x ∈ N

When, y = 1, we get:

x = 8 – 2(1) = 8 – 2

x = 6

When, y = 2, we get

x = 8 – 2(2) = 8 – 4

x = 4

When, y = 3, we get

x = 8 – 2(3) = 8 – 6

x = 2

When, y = 4, we get

x = 8 – 2(4) = 8 – 8

x = 0

Now, y cannot have the value 4 because when y = 4, x becomes zero which is clearly not a natural number.

Therefore, we can write:

R\text{ }=\text{ }\left\{ \left( 2,\text{ }3 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 6,\text{ }1 \right) \right\}

{{R}^{-1}}~=\text{ }\left\{ \left( 3,\text{ }2 \right),\text{ }\left( 2,\text{ }4 \right),\text{ }\left( 1,\text{ }6 \right) \right\}

i

More Material