Question

Let a relation R be defined by {{R}^{-1}}=\left\\{ \left( 5,4 \right),\left( 4,1 \right),\left( 6,4 \right),\left( 6,7 \right),\left( 7,3 \right) \right\\}.
Find (i) $RoR$ (ii) ${{R}^{-1}}oR$

Answer 1

We start solving the problem by recalling the definitions of relations and the general form of the composite relation. We then find the terms that were in the form of $\left( x,y \right)$ in first relation and $\left( y,z \right)$ in second relation to get the terms of the composite relation $R\circ R$. We then follow a similar procedure to find the sets of elements in the composite relation ${{R}^{-1}}\circ R$.

Complete step by step answer:
Let A, B, and C be sets, and let R be a relation from A to B and let S be a relation from B to C. That is, R is a subset of A $\times$ B and S is a subset of B $\times$ C. Then R and S give rise to a relation from A to C indicated by $RoS$. This relation is known as composition of relations. Let R is a relation on a set A, that is, R is a relation from a set A to itself. Then $RoR$, the composition of R with itself, is always represented.
We know that in order to find the composition of relations, we need to follow the condition mentioned below.
If there are two relations R and S such that relation R is between a and b, relation S is between b and c.
\Rightarrow R\circ S=\left\\{ \left( a,c \right)|\text{there exists b} \in \text{B for which} \left( a,b \right)\in R \text{ and} \left( b,c \right)\in S \right\\}.
(i) To obtain the elements of $RoR$, we proceed as follows. We find the pairs from the relation R in such a way that $\left( x,y \right)$ and $\left( y,z \right)$ are obtained.
So, we get such ordered pairs in the relation R as
$\left( i \right)\left( 1,4 \right),\left( 4,5 \right)$,
$\left( ii \right)\left( 1,4 \right),\left( 4,6 \right)$,
$\left( iii \right)\left( 3,7 \right),\left( 7,6 \right)$.
\Rightarrow RoR=\left\\{ \left( 1,5 \right),\left( 1,6 \right),\left( 3,6 \right) \right\\}.
(ii) We first find ${{R}^{-1}}$.
We know that in order to find the inverse of a relation, the following condition has to be followed.
\Rightarrow {{R}^{-1}}=\left\\{ \left( y,x \right)|\left( x,y \right)\in R \right\\}.
We have {{R}^{-1}}=\left\\{ \left( 5,4 \right),\left( 4,1 \right),\left( 6,4 \right),\left( 6,7 \right),\left( 7,3 \right) \right\\}
We now obtain the elements of ${{R}^{-1}}oR$. We first pick the ordered pairs from ${{R}^{-1}}$ and then from R in such a way that $\left( x,y \right)$ present in ${{R}^{-1}}$ and $\left( y,z \right)$ present in R.
We get such ordered pairs as
$\left( i \right)\left( 5,4 \right),\left( 4,5 \right)$,
$\left( ii \right)\left( 5,4 \right),\left( 4,6 \right)$,
$\left( iii \right)\left( 4,1 \right),\left( 1,4 \right)$,
$\left( iv \right)\left( 6,4 \right),\left( 4,5 \right)$,
$\left( v \right)\left( 6,4 \right),\left( 4,6 \right)$,
$\left( vi \right)\left( 6,7 \right),\left( 7,6 \right)$,
$\left( vii \right)\left( 7,3 \right),\left( 3,7 \right)$.
Hence, we get
{{R}^{-1}}oR=\left\\{ \left( 5,5 \right),\left( 5,6 \right),\left( 4,4 \right),\left( 6,5 \right)\left( 6,6 \right),\left( 6,6 \right),\left( 7,7 \right) \right\\}.
{{R}^{-1}}oR=\left\\{ \left( 5,5 \right),\left( 5,6 \right),\left( 4,4 \right),\left( 6,5 \right)\left( 6,6 \right),\left( 7,7 \right) \right\\}.
This is the required solution.

Note: We should know that the first element of $\left( x,z \right)$ will come from the relation ${{R}^{-1}}$ and second element of $\left( x,z \right)$ will come from the relation $R$ while find the composite relation ${{R}^{-1}}\circ R$. If we get the same set more than once in the relationship we neglect one of them. We can solve for the composite relation using the matrices and functions.