we need to check if $ R $ is reflexive, symmetric, and transitive.

Reflexivity:

A relation $ R $ is reflexive if every element is related to itself, i.e., $(A, A) \in R$ for all $ A \in M $.

Given any matrix $ A \in M $, we have

$\operatorname{det}(A-A) = \operatorname{det}(0) = 0$.

Since $ 0 $ is an integer, $(A, A) \in R$ for all $ A \in M$. Therefore, $ R $ is reflexive.

Symmetry:

A relation $ R $ is symmetric if $(A, B) \in R$ implies $(B, A) \in R$.

Suppose $(A, B) \in R$, meaning $\operatorname{det}(A-B)$ is an integer. We need to verify that $\operatorname{det}(B-A)$ is also an integer.

Since

$ \operatorname{det}(B-A) = -\operatorname{det}(A-B), $

and the negative of an integer is also an integer, $\operatorname{det}(B-A)$ is an integer. Thus, $ R $ is symmetric.

Transitivity:

A relation $ R $ is transitive if whenever $(A, B) \in R$ and $(B, C) \in R$, it follows that $(A, C) \in R$.

Let's consider:

$ \begin{align*} A &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \\ B &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ C &= \begin{bmatrix} 0.5 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*} $

We find:

$ \operatorname{det}(A-B) = 1, \quad \operatorname{det}(B-C) = 0, \quad \operatorname{det}(A-C) = 0.5. $

Here, $(A, B) \in R$ and $(B, C) \in R$, but $(A, C) \notin R$ because $ \operatorname{det}(A-C) = 0.5$, which is not an integer. Hence, $ R $ is not transitive.

Based on this analysis, $ R $ is reflexive and symmetric, but not transitive.

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