Here, R₁, R₂, R_3, and R₄ are the binary relations.

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.

We have

R₃ on R defined by (a, b) ∈ R₃⇔ a² – 4ab + 3b² = 0

Check for Reflexivity:

∀ a ∈ R,

(a, a) ∈ R₃ needs to be proved for reflexivity.

If (a, b) ∈ R₃, then we have

a² – 4ab + 3b² = 0

Replace b by a, we get

a² – 4aa + 3a² = 0

⇒ a² – 4a² + 3a² = 0

⇒ –3a² + 3a² = 0

⇒ 0 = 0, which is true.

⇒ (a, a) ∈ R₃

So, ∀ a ∈ R, (a, a) ∈ R₃

∴ R₃ is reflexive.

Check for Symmetry:

∀ a, b ∈ R

If (a, b) ∈ R₃, then we have

a² – 4ab + 3b² = 0

⇒ a² – 3ab – ab + 3b² = 0

⇒ a (a – 3b) – b (a – 3b) = 0

⇒ (a – b) (a – 3b) = 0

⇒ (a – b) = 0 or (a – 3b) = 0

⇒ a = b or a = 3b …(1)

Replace a by b and b by a in equation (1), we get

⇒ b = a or b = 3a …(2)

Results (1) and (2) does not match.

⇒ (b, a) ∉ R₃

∴ R₃ is not symmetric.

Check for Transitivity:

∀ a, b, c ∈ R

If (a, b) ∈ R₃ and (b, c) ∈ R₃

⇒ a² – 4ab + 3b² = 0 and b² – 4bc + 3c² = 0

⇒ a² – 3ab – ab + 3b² = 0 and b² – 3bc – bc + 3c²

⇒ a (a – 3b) – b (a – 3b) = 0 and b (b – 3c) – c (b – 3c) = 0

⇒ (a – b) (a – 3b) = 0 and (b – c) (b – 3c) = 0

⇒ (a – b) = 0 or (a – 3b) = 0

And (b – c) = 0 or (b – 3c) = 0

⇒ a = b or a = 3b And b = c or b = 3c

What we need to prove here is that, a = c or a = 3c

Take a = b and b = c

Clearly implies that a = c.

[∵ if a = b, just substitute a in place of b in b = c. We get, a = c]

Now, take a = 3b and b = 3c

If a = 3b

b = a/3

Substitute b = a/3 in b = 3c. We get

a/3 = 3c

⇒ a = 9c, which is not the desired result.

⇒ (a, c) ∉ R₃

∴ R₃ is not transitive.