Step 1: Understanding the Concept:
A mathematical relation $R$ defined on a set $A$ can be classified into different categories by checking how elements pair with each other. The three primary properties are:
1. Reflexive: Every element must map to itself. For every $a \in A$, the ordered pair $(a, a)$ must be in $R$.
2. Symmetric: If an element maps to another, the reverse must also be true. If $(a, b) \in R$, then $(b, a)$ must be in $R$.
3. Transitive: If a chain link exists between three elements, a direct link must join the endpoints. If $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must be in $R$.
Step 2: Detailed Explanation:
Let's test our set $A = \{1, 2, 3\}$ against the pairs in relation $R = \{(1,1), (2,2), (3,3), (1,2)\}$ to see which rules hold true:
- Check Reflexivity: Our base set $A$ contains elements 1, 2, and 3. For $R$ to be reflexive, it must contain pairs $(1,1)$, $(2,2)$, and $(3,3)$. Looking at our relation set, all three identity pairs are present. Therefore, R is Reflexive.
- Check Symmetry: The relation contains the ordered pair $(1,2)$. For $R$ to be symmetric, the flipped pair $(2,1)$ must also be present in the set. Checking our relation components, $(2,1) \notin R$. Therefore, R is NOT Symmetric. (This immediately rules out options A and C).
- Check Transitivity: Let's look for connected pairs where the second element of one matches the first element of another:
- Take pairs $(1,1)$ and $(1,2) \implies$ the shortcut pair is $(1,2)$, which is in $R$.
- Take pairs $(1,2)$ and $(2,2) \implies$ the shortcut pair is $(1,2)$, which is in $R$.
No conflicting broken chain links exist anywhere in this set. Therefore, R is Transitive.
Since the relation is both reflexive and transitive but not symmetric, it satisfies option (B).
Step 3: Final Answer:
The properties satisfied by R are Reflexive and Transitive.