Question:
If $R$ s a relation defined on set $A=\{1,2,3\}$ by the rule $\left(a_{t} b\right) \in R \Leftrightarrow\left|a^{2}-b^{2}\right| \leq 5$, then $R^{-1}=$
Solution:
Given: $R=\left\{(a, b):\left|a^{2}-b^{2}\right| \leq 5\right\}$, where $A=\{1,2,3\}$ and $a, b \in A$.
R = {(1, 1), (1, 2), (2, 1), (2, 2), (2, 3), (3, 2), (3, 3)}
Therefore, $R^{-1}=\{(1,1),(2,1),(1,2),(2,2),(3,2),(2,3),(3,3)\}=R$
Hence, if $R$ is a relation defined on set $A=\{1,2,3\}$ by the rule $(a, b) \in R \Leftrightarrow\left|a^{2}-b^{2}\right| \leq 5$, then $R^{-1}=\underline{R}$.