Consider the following relation R on the set of real square matrices of order 3 . $R=\left\{(A, B) \mid A=P^{-1} B P\right.$ for some invertible matrix $\left.P\right\}$.

Question

 Consider the following relation $\mathrm{R}$ on the set of real square matrices of order 3 .

$R=\left\{(A, B) \mid A=P^{-1} B P\right.$ for some invertible matrix $\left.P\right\}$.

Statement -1 : $R$ is equivalence relation.

Statement - 2 : For any two invertible $3 \times 3$ matrices $M$ and $N,(M N)^{-1}=N^{-1} M^{-1}$.

(1) Statement- 1 is true, statement- 2 is a correct explanation for statement- 1 .

(2) Statement- 1 is true, statement- 2 is true; statement- 2 is not a correct explanation for statement-1.

(3) Statement- 1 is true, statement-2 is false.

(4) Statement- 1 is false, statement- 2 is true.

Answer 1

SOLUTION : or reflexive

$(A, A) \in R$

$\Rightarrow \quad A=P^{-1} A P$

which is true for $P=1$

$\therefore$ reflexive

for symmetry

As $(A, B) \in R$ for matrix $P$

$A=P^{-1} B P \\$

$\Rightarrow \quad \mathrm{PA}=\mathrm{PP}^{-1} \mathrm{BP} \\$

$\Rightarrow \quad \mathrm{PAP}^{-1}=\mathrm{IBPP}^{-1} \\$

$\Rightarrow \quad \mathrm{PAP}^{-1}=|\mathrm{B}| \\$

$\Rightarrow \quad \mathrm{PAP}^{-1}=\mathrm{B} \\$

$\Rightarrow \quad B=\text { PAP }^{-1} \\$

$\therefore \quad(B, A) \in R \text { for matrix } P^{-1} \\$

$\therefore \quad \mathrm{R} \text { is symmetric } \\$

for transitivity

$A=P^{-1} B P$

and $B=P^{-1} C P$

$\Rightarrow  A=P^{-1}\left(P^{-1} C P\right) P \\$

$\Rightarrow  A=\left(P^{-1}\right)^2 C P^2 \\$

$\Rightarrow A=\left(P^2\right)^{-1} C\left(P^2\right) \\$

$\therefore (A, C) \in R \text { for matrix } P^2 \\$

$\therefore  \text { R is transitive }$

$\therefore \quad(A, C) \in R$ for matrix $P^2$

$\therefore \quad \mathrm{R}$ is transitive

so $R$ is quivalenceginal image