Question

11. Let A=QxQ.Let * be a binary operation on A defined by : (a,b)*(c,d) = (ac,ad+b).Find i) Identity element of (A,*) ii) the invertible element of (A,*).

Answer 1

(a,b) * (c,d) = (ac, ad+b)

i) Identity Element. 
Let (e,f) be the identity element. 
Then, (a,b) * (e,f) = (ae, af+b) = (a,b) [a*e = a, where e is the identity element]
(ae,af+b) = (a,b)
⇒ ae = a and af+b = b
⇒ e = 1 and f = 0. 
(1,0) is the identity element

ii) Invertible element 
 Let (c,d) be the inverse of (a,b)
then, 
(a,b) * (c,d) = (1,0) [a*b = e, e is the identity]

⇒ (ac, ad+b) = (1,0)
⇒ ac = 1 and ad+b = 0
⇒ c = 1/a and d = -b/a
therefore, (1/a, -b/a) is the inverse.

Answer 2

A is defined on rational numbers  
A = QxQ 

So (a,b)*(c,d) = (ac,ad+b) 

 

1) for identifying element we have a*e = a 

Let a = (a1,a2) and e = (e1,e2) 

So a*e = (a1e1, a1e2 + a2) 

This should be equal to a = (a1,a2) 

So e = (1,0) 

(a1e1, a1e2 + a2) = (a1(1), a1(0) + a2 = (a1,a2) 

 

2) for invertible element we have a*b=e=b*a 

We have found e already  

So as a = (a,b) , and b = (x1,x2) 

So a*b = (ax1,ax2 + b) = (1,0) 

x1 = 1/a and x2 = -b/a 

So the invertible element is (1/a, -b/a)