Symmetric relation

Step 1: Exactly two diagonal pairs can be selected in (\binom{4}{2}=6) ways. Step 2: There are (4\cdot3/2=6) unordered off-diagonal pair groups, giving (2^6) choices. Step 3: Multiply the diagonal choices and off-diagonal symmetric choices.

Step 1: All three given pairs are off-diagonal. Step 2: Symmetry forces ((2,1),(3,2),(4,3)), making (6) off-diagonal pairs in total. Step 3: For a minimum count, do not add extra diagonal or unrelated pairs.

Step 1: For a symmetric relation, (R^{-1}=R). Step 2: Therefore, (R\cup R^{-1}=R\cup R=R). Step 3: In inverse-relation questions, identify (R^{-1}) first.

The governing idea is symmetric closure. The inverse relation R⁻¹ contains the reverse pair for every pair in R. Let S = R ∪ R⁻¹. If (a,b) belongs to S, then either (a,b) belongs to R or it belongs to R⁻¹. In the first case, (b,a) belongs to R⁻¹; in the second case, (b,a) belongs to R. Thus in both cases the reverse pair belongs to S, proving that S is symmetric. The union need not be reflexive because R may contain no self-pairs, and it need not be empty. Since the union of two relations on the same sets is itself a relation, option D is also false. Hence A is always true.

Step 1: To make (R) symmetric, every pair needs its reverse. Step 2: (R^{-1}) provides exactly those reverse pairs, so (R\cup R^{-1}) contains all required pairs. Step 3: When the word smallest appears, reject options that add unnecessary pairs.

Step 1: (R\cap R^{-1}=R) implies (R\subseteq R^{-1}). Step 2: This means every pair of (R) has its reverse in (R), so (R) is symmetric. Step 3: Inclusion statements such as (R\subseteq R^{-1}) are useful for proving symmetry.

Step 1: In a relation matrix, symmetry requires (m_{ij}=m_{ji}). Step 2: Here (m_{12}=0) but (m_{21}=1), so the relation is not symmetric. Step 3: One mismatched pair of matrix entries is enough to break symmetry.

Step 1: In a symmetric relation, the number of off-diagonal pairs is always even because they appear in reverse pairs. Step 2: If there are (3) diagonal pairs out of (10), the off-diagonal count is (7), which is not even. Step 3: Checking parity of total and diagonal counts is a fast exam method.

Step 1: Swapping (a) and (b) in (a^2+b^2=25) gives (b^2+a^2=25). Step 2: The value does not change, so the reverse pair also belongs to the relation. Step 3: Rules based on sums of squares are often directly symmetric.

Step 1: (a^2-b^2=0) means (a^2=b^2). Step 2: If (a^2=b^2), then (b^2=a^2), so ((b,a)) also belongs to the relation. Step 3: Equality remains true when the two sides are interchanged.

Step 1: For ((3,1)), (3=1+2) is true, so the pair belongs to the relation. Step 2: For the reverse ((1,3)), (1=3+2) is false. Step 3: In direction-dependent rules, always test the reverse pair separately.

Step 1: The (n) diagonal pairs can be chosen independently. Step 2: Off-diagonal pairs are chosen in reverse-pair groups, and there are (\frac{n(n-1)}{2}) such groups. Step 3: The total independent choices are (n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}).

Step 1: In a symmetric relation, the number of off-diagonal pairs is even. Step 2: Since the total is (5), the number of diagonal pairs must be odd. With (3) elements, possible diagonal counts are at most (3), so (1) is possible. Step 3: Apply parity first, then check the maximum limit.

Step 1: Since (R) is symmetric, (R^{-1}=R). Step 2: Therefore (R\cup R^{-1}=R), and (R\subseteq T\subseteq R) gives (T=R). Step 3: In inclusion-based questions, simplify the relation expressions first.

Let S=(A×A)−R be the complement of R in A×A. To test symmetry, assume (a,b)∈S, so (a,b)∉R. If (b,a) were in R, then symmetry of R would imply (a,b)∈R, contradicting the assumption. Therefore (b,a)∉R and hence (b,a)∈S. Thus S is symmetric. The complement need not be empty or diagonal-only, and it can be symmetric even when R is neither universal nor empty.

Step 1: One off-diagonal pair group means ((a,b)) and ((b,a)) together. Step 2: Therefore one group gives two ordered pairs. Step 3: In symmetric relations, count off-diagonal selections as groups, not single pairs.

Step 1: In a symmetric relation, ((4,1)\in R) would force ((1,4)\in R). Step 2: Since ((1,4)\notin R), ((4,1)) cannot be in (R). Step 3: In symmetric relations, absence of an off-diagonal pair also implies absence of its reverse.

Step 1: (a\equiv b \pmod{3}) means (a) and (b) have the same remainder on division by (3). Step 2: If (a) and (b) have the same remainder, then (b) and (a) also have the same remainder. Step 3: For same-remainder relations, check the reversed order mentally.

Step 1: Test one pair for symmetry. ((2,1)) belongs because (2\equiv 2\cdot1 \pmod{5}). Step 2: The reverse ((1,2)) would require (1\equiv4 \pmod{5}), which is false. Step 3: Modular relations with a multiplier do not always survive reversal.

Step 1: There are (3) ways to choose one diagonal pair. Step 2: Two off-diagonal ordered pairs mean one reverse-pair group; with (3) elements, there are (\binom{3}{2}=3) such groups. Step 3: Total possibilities are (3\cdot3=9).