We have been given,

A = {1, 2, 3}

Here, R₁, R_2, and R₃ 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 R₁.

R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∈ R₁

(2, 2) ∈ R₂

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∈ R₁

∴ R₁ is reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁, then (3, 1) ∈ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But if (2, 1) ∈ R₁, then (1, 2) ∉ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

So, if (a, b) ∈ R₁, then (b, a) ∉ R₁

∀ a, b ∈ A

∴ R₁ is not symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₁ and (3, 3) ∈ R₁

Then, (1, 3) ∈ R₁

[∵ R₁ = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}]

But, if (2, 1) ∈ R₁ and (1, 3) ∈ R₁

Then, (2, 3) ∉ R₁

So, if (a, b) ∈ R₁ and (b, c) ∈ R₁, then (a, c) ∉ R₁

∀ a, b, c ∈ A

∴ R₁ is not transitive.

Now, take R₂.

R₂ = {(2, 2), (3, 1), (1, 3)}

(i). Reflexive:

∀ 1, 2, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₂

(2, 2) ∈ R₂

(3, 3) ∉ R₂

So, for a ∈ A, (a, a) ∉ R₂

∴ R₂ is not reflexive.

(ii). Symmetric:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂, then (3, 1) ∈ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

If (2, 2) ∈ R₂, then (2, 2) ∈ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂, then (b, a) ∈ R₂

∀ a, b ∈ A

∴ R₂ is symmetric.

(iii). Transitivity:

∀ 1, 2, 3 ∈ A

If (1, 3) ∈ R₂ and (3, 1) ∈ R₂

Then, (1, 1) ∉ R₂

[∵ R₂ = {(2, 2), (3, 1), (1, 3)}]

So, if (a, b) ∈ R₂ and (b, c) ∈ R₂, then (a, c) ∉ R₂

∀ a, b, c ∈ A

∴ R₂ is not transitive.

Now take R₃.

R₃ = {(1, 3), (3, 3)}

(i). Reflexive:

∀ 1, 3 ∈ A [∵ A = {1, 2, 3}]

(1, 1) ∉ R₃

(3, 3) ∈ R₃

So, for a ∈ A, (a, a) ∉ R₃

∴ R₃ is not reflexive.

(ii). Symmetric:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃, then (3, 1) ∉ R₃

[∵ R₃ = {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃, then (b, a) ∉ R₃

∀ a, b ∈ A

∴ R₃ is not symmetric.

(iii). Transitivity:

∀ 1, 3 ∈ A

If (1, 3) ∈ R₃ and (3, 3) ∈ R₃

Then, (1, 3) ∈ R₃

[∵ R₃ = {(1, 3), (3, 3)}]

So, if (a, b) ∈ R₃ and (b, c) ∈ R₃, then (a, c) ∈ R₃

∀ a, b, c ∈ A

∴ R₃ is transitive.