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}
iii. R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3
An inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a).
i. Given, R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
∴ R‑1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}
⇒ R‑1 = {(2, 1), (2, 3), (3, 1), (3, 2), (6, 5)}
ii. Given, R= {(x, y) : x, y N; x + 2y = 8}
Here, x + 2y = 8
⇒ x = 8 – 2y
As y N, Put the values of y = 1, 2, 3,…… till x N
On putting y=1, x = 8 – 2(1) = 8 – 2 = 6
On putting y=2, x = 8 – 2(2) = 8 – 4 = 4
On putting y=3, x = 8 – 2(3) = 8 – 6 = 2
On putting y=4, x = 8 – 2(4) = 8 – 8 = 0
Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.
∴ R = {(2, 3), (4, 2), (6, 1)}
R‑1 = {(3, 2), (2, 4), (1, 6)}
⇒ R‑1 = {(1, 6), (2, 4), (3, 2)}
iii. Given, R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3
Here,
x {11, 12, 13} and y (8, 10, 12}
y = x – 3
On putting x = 11, y = 11 – 3 = 8 (8, 10, 12}
On putting x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}
On putting x = 13, y = 13 – 3 = 10 (8, 10, 12}
∴ R = {(11, 8), (13, 10)}
R‑1 = {(8, 11), (10, 13)}