Question

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) $R=\left\{\right(x,y):x,yϵN,x+2y=8\}$

(iii) R is a relation form {11, 12, 13} to {8, 10, 12} defined by y = x - 3.

Answer 1

(i) We have,

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

(ii) We have,

$R=\left\{\right(x,y):x,yϵN,x+2y=8\}$

Now,

$x+2y=8\Rightarrow x=8-2y$

Putting y = 1, 2, 3 we get x = 6, 4, 2 respectively,

For y = 4, we get $x=0\text{/}ϵN.$ Also for $y>4\text{/}ϵN$

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

Thus,

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

$\Rightarrow R^{-1}\left\{\right(3,2),(2,4),(1,6\left)\right\}$

(iii) We have,

R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x - 3

Now,

y = x - 3

Putting x = 11, 12, 13 we get y = 8, 9, 10 respetively

$\Rightarrow (11,8)\text{/}ϵR,(12,9)\text{/}ϵR$ and $(13,10)ϵR$

Thus,

R = {(11, 8), (13, 10)}

$\Rightarrow R^{-1}=\left\{\right(8,11),(10,13\left)\right\}$