Answer :

We have been given,

A = {1, 2, 3}

Here, R_{1}, R_{2,} and R_{3} are the binary relations on A.

So, recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

So, using these results let us start determining given relations.

Let us take R_{1}.

R_{1} = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

(i). **Reflexive:**

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∈ R_{1}

(2, 2) ∈ R_{2}

(3, 3) ∈ R_{3}

So, for a ∈ A, (a, a) ∈ R_{1}

**∴** **R _{1} is reflexive.**

(ii). **Symmetric:**

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R_{1}, then (3, 1) ∈ R_{1}

[∵ R_{1} = {(1, 1), **(1, 3), (3, 1)**, (2, 2), (2, 1), (3, 3)}]

But if (2, 1) ∈ R_{1}, then (1, 2) ∉ R_{1}

[∵ R_{1} = {(1, 1), (1, 3), (3, 1), (2, 2), **(2, 1)**, (3, 3)}]

So, if (a, b) ∈ R_{1}, then (b, a) ∉ R_{1}

∀ a, b ∈ A

**∴** **R _{1} is not symmetric.**

(iii). **Transitivity:**

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R_{1} and (3, 3) ∈ R_{1}

Then, (1, 3) ∈ R_{1}

[∵ R_{1} = {(1, 1), **(1, 3)**, (3, 1), (2, 2), (2, 1), **(3, 3)**}]

But, if (2, 1) ∈ R_{1} and (1, 3) ∈ R_{1}

Then, (2, 3) ∉ R_{1}

So, if (a, b) ∈ R_{1} and (b, c) ∈ R_{1}, then (a, c) ∉ R_{1}

∀ a, b, c ∈ A

**∴** **R _{1} is not transitive.**

Now, take R_{2}.

R_{2} = {(2, 2), (3, 1), (1, 3)}

(i). **Reflexive:**

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R_{2}

(2, 2) ∈ R_{2}

(3, 3) ∉ R_{2}

So, for a ∈ A, (a, a) ∉ R_{2}

**∴** **R _{2} is not reflexive.**

(ii). **Symmetric:**

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R_{2}, then (3, 1) ∈ R_{2}

[∵ R_{2} = {(2, 2), **(3, 1), (1, 3)**}]

If (2, 2) ∈ R_{2}, then (2, 2) ∈ R_{2}

[∵ R_{2} = {**(2, 2),** (3, 1), (1, 3)}]

So, if (a, b) ∈ R_{2}, then (b, a) ∈ R_{2}

∀ a, b ∈ A

**∴** **R _{2} is symmetric.**

(iii). **Transitivity:**

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R_{2} and (3, 1) ∈ R_{2}

Then, (1, 1) ∉ R_{2}

[∵ R_{2} = {(2, 2), **(3, 1), (1, 3)**}]

So, if (a, b) ∈ R_{2} and (b, c) ∈ R_{2}, then (a, c) ∉ R_{2}

∀ a, b, c ∈ A

**∴** **R _{2} is not transitive.**

Now take R_{3}.

R_{3} = {(1, 3), (3, 3)}

(i). **Reflexive:**

∀ 1, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R_{3}

(3, 3) ∈ R_{3}

So, for a ∈ A, (a, a) ∉ R_{3}

**∴** **R _{3} is not reflexive.**

(ii). **Symmetric:**

∀ 1, 3 ∈ A

If (1, 3) ∈ R_{3}, then (3, 1) ∉ R_{3}

[∵ R_{3} = {**(1, 3)**, (3, 3)}]

So, if (a, b) ∈ R_{3}, then (b, a) ∉ R_{3}

∀ a, b ∈ A

**∴** **R _{3} is not symmetric.**

(iii). **Transitivity:**

∀ 1, 3 ∈ A

If (1, 3) ∈ R_{3} and (3, 3) ∈ R_{3}

Then, (1, 3) ∈ R_{3}

[∵ R_{3} = {**(1, 3), (3, 3)**}]

So, if (a, b) ∈ R_{3} and (b, c) ∈ R_{3}, then (a, c) ∈ R_{3}

∀ a, b, c ∈ A

**∴** **R _{3} is transitive.**

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