Question

Let M be the set of all $3\times 3$matrices with elements integers. A relation R on M is defined such that ARB if (A – B) is skew – symmetric. Then the relation R is,
A. An equivalence relation
B. Reflexive, symmetric, but not transitive
C. Reflexive, not symmetric, not transitive
D. Symmetric and transitive, but not reflexive

Answer 1

Hint: We will be using the concepts of functions and relations to solve the problem. We will be using the definitions of reflexive relation, symmetric relations and transitive relations to verify if each relation holds or not and hence deduce the answer.

Complete step-by-by answer:

Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive or a combination of these.
Now, we know that reflexive relations are those in which every element is mapped to itself i.e. $\left( a,a \right)\in R$ while symmetric relations are those for which if a R b then b R a. Also, holds and transitive are those relations in which if a R b and b R c then a R c must be held.
Now, we know different types of relations. We will check the given relation for these.
Now, we have been given a relation R on m such that ARB iff (A – B) is skew – symmetric.
Now, we know a skew symmetric matrix A is such that ${{A}^{T}}=-A$.
Now, for reflexive we have ARA as (A – A) is skew symmetric which is true. Since (A – A) is a zero matrix which is a skew symmetric always.
Now, we have for symmetric that if $ARB\Rightarrow \left( A-B \right)$ is skew symmetric. Therefore,
$\begin{align}
  & {{\left( A-B \right)}^{T}}=-\left( A-B \right) \\
 & {{A}^{T}}-{{B}^{T}}=-A+B \\
\end{align}$
Now, multiply by -1 on both sides,
$\begin{align}
  & {{B}^{T}}-{{A}^{T}}=A-B \\
 & {{B}^{T}}-{{A}^{T}}=-\left( B-A \right) \\
\end{align}$
So, we have BRA. Therefore, $ARB\Rightarrow BRA$ hence, R is symmetric also.
Now, we have for transitive, that if
$\begin{align}
  & ARB\Rightarrow {{\left( A-B \right)}^{T}}=-\left( A-B \right)..........\left( 1 \right) \\
 & BRC\Rightarrow {{\left( B-C \right)}^{T}}=-\left( B-C \right)..........\left( 2 \right) \\
\end{align}$
Now, adding (1) and (2) we have,
$\begin{align}
  & {{\left( A-B \right)}^{T}}+{{\left( B-C \right)}^{T}}=-\left( A-B \right)-\left( B-C \right) \\
 & {{A}^{T}}-{{B}^{T}}+{{B}^{T}}-{{C}^{T}}=-A+B-B+C \\
 & {{A}^{T}}-{{C}^{T}}=-A+C \\
 & {{\left( A-C \right)}^{T}}=-\left( A-C \right) \\
\end{align}$
So, we have ARC
Now, since $ARB\ and\ BRC\Rightarrow ARC$. Therefore, the relation is transitive also.
Now, we know that a relation is symmetric, reflexive and transitive is an equivalence relation.
Therefore, the correct answer is (A).

Note: To solve these types of questions it is important to note that a R b means that a is related to b by a relation R. Also these types of questions are solved easily by giving examples and counterexamples. Also, we have to check the relation for reflexive, symmetric and transitive relation to check it for equivalence relation.