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)}


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