(i). Let (x, y) be the identity element in A.
Now,
(a, b) * (x, y) = (a, b)
= (x, y) * (a, b) ∀ (a, b) ∈ A
⇒ (ax, b + ay) = (a, b) = (xa, y + bx)
Equating corresponding terms, we get
⇒ ax = a, b + ay = b or a = xa, b = y + bx,
⇒ x = 1 and y = 0
Hence, (1, 0) is the identity element in A.
(ii). Let (a,b) be an invertible element in A and let (c,d) be its inverse in A.
Now,
(a,b) * (c,d) = (1,0) = (c,d) * (a,b)
⇒ (ac,b+ad) = (1,0) = (ca,d+c)
⇒ ac = 1, b + ad = 0
or 1 = ca, 0 = d + bc [By equating coefficients]
⇒ \(c = \frac{1}{a}\) and \(d=-\frac{b}{a}\)
Where, a ≠ 0
Therefore, all (a,b) ∈ A is an invertible element of A if a ≠ 0, and inverse of (a,b) is (\( \frac{1}{a}\), \(-\frac{b}{a}\))
For inverse of (5,3)
Inverse of (5,3) = (\(\frac{1}{5},\,-\frac{3}{5}\)) (∵ Inverse of (a,b) =\( \frac{1}{a}\),\(-\frac{b}{a}\))
For inverse of (\( \frac{1}{2}\),4)
Inverse of (\( \frac{1}{2}\),4) = (2, -8)