Question

Consider a relation $R$ defined on set of involutory square matrix of order $2$. If $A,B$ are two such matrices then relation $R$ is defined as

$(A,B)\in R\Rightarrow (AB{)}^T=A^T{B}^T$.

Relation $R$ is

• A. a reflexive relation only
• B. reflexive and symmetric relation only
• C. a transitive relation only
• D. an equivalence relation

Answer 1

Answer: (B)

reflexive and symmetric relation only

Given,

$(AB{)}^T=A^T{B}^T$
$\Rightarrow (AB{)}^T=(BA{)}^T$
$\Rightarrow AB=BA$
So, $(A,B)\in R\Rightarrow AB=BA$
(i) Reflexive:
Since, $AA=AA$ is true.
$\Rightarrow (A,A)$ $\in R$
(ii) Symmetric: Let $(A,B)\in R\Rightarrow AB=BA$
or, $BA=AB\Rightarrow (B,A)\in R$
Hence, $R$ is symmetric.

Note: AB are commutative

as $(AB{)}^T=B^T{A}^T$

but we are given $(AB{)}^T=A^T{B}^T$

$B^T{A}^T=A^T{B}^T$

$(AB{)}^T=(BA{)}^T$

$AB=BA$
(iii) Take $A=\left[\begin{array}{cc}2& -3\\ 1& -2\end{array}\right]$, B =$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, C =$\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$,

then $AB=BA$ and $BC=CB$ but $AC\ne CA$

$AC=\left[\begin{array}{cc}2& 3\\ 1& 2\end{array}\right]$, and $C^T{A}^T=\left[\begin{array}{cc}2& -3\\ -1& 2\end{array}\right]$

So not transitive.

Hence, option 'B' is correct.