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.
(i) R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
R−1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}
(ii) R = {(x, y) : x, y ∈ N, x + 2y = 8}
On solving x + 2y = 8, we get:
x = 8
On putting y = 1, we get x = 6.
On putting y = 2, we get x = 4.
On putting y = 3, we get x = 2.
∴ R = {(6, 1), (4, 2), (2, 3)}
Or,
R−1 = {(1, 6), (2, 4), (3, 2)}
(iii) R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x − 3.
x belongs to {11, 12, 13} and y belongs to {8, 10, 12}.
Also, 11 − 3 = 8 and 13 − 3 = 10
∴ R = {(11, 8), (13,10)}
Or,
R−1 = {(8, 11), (10,13)}