a. The expressions for c and d in terms of a and b are
b. The inverse of (2,1) is (13/15, -4/15)
The question has to do with binary operation.
A binary operation is a rule of mathematical operation on two sets of elements defined on a given set.
Since The operation * is defined over the set of ordered pairs by (a,b)*(c,d)=(ac-bd,ad+bc) for all (a,b),(c,d)€R given that the inverse of (a,b) is (c,d).
Let (e, e') be the identity element.
We know that for any binary operation * on a given set, a*e = a where e is the identity element.
So, (a,b)*(e,e') = (a,b)
So, (a,b)*(c,d) = (ac - bd, ad + bc)
(a,b)*(e,e') = (ae - be, ae' + be')
Since (a,b)*(e,e') = (a,b)
(ae - be, ae' + be') = (a, b)
So,
e = a/(a - b) and e' = b/(a + b)
So, (e, e') = (a/(a - b), b/(a + b))
Also, for any binary operation * under a given set a*a⁻¹ = e where a⁻¹ = inverse of a.
Since (c,d) is the inverse of (a,b), we have that
(a,b)*(c,d) = (e, e')
So, (a,b)*(c,d) = (ac - bd, ad + bc)
= (a/(a - b), b/(a + b))
So,
From (3), c = bd/a + 1/(a - b)
Substituting c into (4), we have
ad + bc = b/(a + b)
ad + b[bd/a + 1/(a - b)] = b/(a + b)
ad + b²d/a + b/(a - b) = b/(a + b)
ad + b²d/a = b/(a + b) - b/(a - b)]
(a + b²/a)d = b[1/(a + b) - 1/(a - b)]
[(a² + b²)/a]d = b[a - b - (a + b)]/(a - b)(a + b)
[(a² + b²)/a]d = b[a - b - a - b)]/(a - b)(a + b)
[(a² + b²)/a]d = b(-2b)/(a² - b²)
d = -2ab²/[(a² - b²)(a² + b²)]
d = -2ab²/(a⁴ - b⁴)
Substitutind d into c, we have
c = bd/a + 1/(a - b)
c = -2ab³/(a⁴ - b⁴)a + 1/(a - b)
c = -2b³/(a⁴ - b⁴) + 1/(a - b)
So, the expressions for c and d in terms of a and b are
Since (c,d) is the inverse of (a,b) is
So, with a = 2 and b = 1, c = -2b³/(a⁴ - b⁴) + 1/(a - b)
= -2(1)³/(2⁴ - 1⁴) + 1/(2 - 1)
= -2/(16 - 1) + 1/1
= -2/15 + 1
= (-2 + 15)/15
= 13/15
So, with a = 2 and b = 1, d = -2ab²/(a⁴ - b⁴)
= -2(2)(1)²/(2⁴ - 1⁴)
= -4(1)/(16 - 1)
= -4/15
So, the inverse of (2,1) is (13/15, -4/15)
Learn more about binary operation here:
https://brainly.com/question/16827196
#SPJ1
