Let `A=QxxQ` , where Q is the set of all rational numbers and `'**'` be the operation on A defined by :
`(a,b)**(c,d)=(ac,b+ad)" for "(a,b),(c,d)inA`.
Then, find : (i) The identity element of `'**'` in A
(ii) Invertible elements of A and hence write the inverse of elements (5, 3) and `((1)/(2),4)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the identity element of the operation '**' in A.

The operation is defined as:
\[ (a,b) ** (c,d) = (ac, b + ad) \]
To find the identity element \( e = (e_1, e_2) \) such that for any \( (a,b) \in A \):
\[ (a,b) ** (e_1, e_2) = (a,b) \]
This means:
\[ (ac, b + ad) = (a,b) \]
From the first component, we have:
\[ a e_1 = a \implies e_1 = 1 \text{ (since } a \text{ can be any rational number)} \]
From the second component, we have:
\[ b + a e_2 = b \implies a e_2 = 0 \]
This must hold for all \( a \), which implies:
\[ e_2 = 0 \]
Thus, the identity element is:
\[ e = (1, 0) \]

### Step 2: Find the invertible elements of A.

An element \( (a,b) \) is invertible if there exists \( (c,d) \) such that:
\[ (a,b) ** (c,d) = (1,0) \]
This gives us the equations:
1. \( ac = 1 \)
2. \( b + ad = 0 \)

From the first equation, we can express \( c \):
\[ c = \frac{1}{a} \quad (\text{provided } a \neq 0) \]
From the second equation, substituting \( c \):
\[ b + a \left(-\frac{b}{a}\right) = 0 \implies b - b = 0 \]
Thus, we find:
\[ d = -\frac{b}{a} \]
So the inverse of \( (a,b) \) is:
\[ \left( \frac{1}{a}, -\frac{b}{a} \right) \]

### Step 3: Find the inverses of the specific elements \( (5,3) \) and \( \left( \frac{1}{2}, 4 \right) \).

1. For \( (5,3) \):
- \( a = 5 \), \( b = 3 \)
- The inverse is:
\[ \left( \frac{1}{5}, -\frac{3}{5} \right) \]

2. For \( \left( \frac{1}{2}, 4 \right) \):
- \( a = \frac{1}{2} \), \( b = 4 \)
- The inverse is:
\[ \left( 2, -8 \right) \]

### Final Answers:
- (i) The identity element is \( (1, 0) \).
- (ii) The inverses are:
- For \( (5, 3) \): \( \left( \frac{1}{5}, -\frac{3}{5} \right) \)
- For \( \left( \frac{1}{2}, 4 \right) \): \( (2, -8) \)