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?
Best Answer
You're correct. Since the definition of the given relation uses the equality relation (which is itself reflexive, symmetric, and transitive), we get that the given relation is also reflexive, symmetric, and transitive pretty much for free.
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 $X\ne Y$.