R is a relation from {11, 12, 13} to {8, 10, | vedclass

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(A) Given the relation $R$ from set $A = \{11, 12, 13\}$ to set $B = \{8, 10, 12\}$ defined by $y = x - 3$.We check the elements of $A$ to see which s

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$\{(8, 11), (10, 13)\}$

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(A) Given the relation $R$ from set $A = \{11, 12, 13\}$ to set $B = \{8, 10, 12\}$ defined by $y = x - 3$.We check the elements of $A$ to see which satisfy the condition $y = x - 3$ for $y \in B$:For $x = 11$,$y = 11 - 3 = 8$. Since $8 \in B$,$(11, 8) \in R$.For $x = 12$,$y = 12 - 3 = 9$. Since $9 
otin B$,$(12, 9) 
otin R$.For $x = 13$,$y = 13 - 3 = 10$. Since $10 \in B$,$(13, 10) \in R$.Thus,$R = \{(11, 8), (13, 10)\}$.The inverse relation ${R^{-1}}$ is obtained by interchanging the elements of the ordered pairs in $R$.Therefore,${R^{-1}} = \{(8, 11), (10, 13)\}$.

$R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3$. Then ${R^{ - 1}}$ is

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