# Symmetric, transitive and reflexive properties of a matrix

Say I had a relation \begin{align} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align} where $a,b,c,d \in \mathbb{R}$, where $X$ is related to $Y$ if and only if $\det(X) = \det(Y)$ (where $\det(A) = ad-bc$)

So I say it is reflexive since $xRx$ since $\det(X) = \det(X)$

However would it be Symmetric: $xRy$ implies $yRx$ (I would say yes, since to be related $\det(x)=\det(Y)$ so therefore $\det(y) = \det(x)$?)

Transitivity is a hard, $xRy yRb$ then $xRb$ (I would say yes since if $xRy$, $\det(x)=\det(y)$ and then $\det(y) = \det(b)$, so $xRb$ since $\det(b) = \det(x)$

And not anti symmetric since if $X=\{1, 2, 3, 4\}$ and $Y = \{4, 2, 3, 1\}$ then they would be $xRy$ but $X$ does not equal $Y$.

Am I on the right track with these?

To show that the given relation is not antisymmetric, your counterexample is correct. If we choose matrices $X,Y\in \left\{ \left[ \begin{array}{cc} a & b \\ c & d \end{array}\right] \text{ } \middle| \text{ } a,b,c,d\in \mathbb{R}\right\}$ , where: $$X=\left[ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right] \text{ and } Y=\left[ \begin{array}{cc} 4 & 2 \\ 3 & 1 \end{array}\right]$$
Then certainly $X$ is related to $Y$ since $\det(X)=1\cdot4-2\cdot3=-2=4\cdot1-2\cdot3=\det(Y)$. Likewise, since the relation was proven to be symmetric, we know that $Y$ is related to $X$. Yet $X\ne Y$.