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

Hint: 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.