Let $A = \{1, 2, 3\}, B = \{1, 3, 5\}$. $A$ relation $R:A \to B$ is defined by $R = \{(1, 3), (1, 5), (2, 1)\}$. Then ${R^{ - 1}}$ is defined by
$\{(1,2), (3,1), (1,3), (1,5)\}$
$\{(1, 2), (3, 1), (2, 1)\}$
$\{(1, 2), (5, 1), (3, 1)\}$
None of these
Let $H$ be the set of all houses in a village where each house is faced in one of the directions, East, West, North, South. Let $R = \{ (x,y)|(x,y) \in H \times H$ and $x, y$ are faced in same direction $\}$ . Then the relation $' R '$ is
Let $R_{1}$ and $R_{2}$ be two relations defined on $R$ by $a R _{1} b \Leftrightarrow a b \geq 0$ and $a R_{2} b \Leftrightarrow a \geq b$, then
Let $R$ be a relation on $Z \times Z$ defined by$ (a, b)$$R(c, d)$ if and only if $ad - bc$ is divisible by $5$ . Then $\mathrm{R}$ is
Let $R$ be a relation on a set $A$ such that $R = {R^{ - 1}}$, then $R$ is
Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.