Question -
Answer -
(i) Let X = (a, b) andY = (c, d) ∈ A, ∀ a, c ∈ R0 and b, d ∈ R
Then, X O Y = (ac, bc+ d)
And Y O X = (ca, da +b)
Therefore,
X O Y = Y O X, ∀ X, Y ∈ A
Thus, O is not commutativeon A.
Now we have to checkassociativity of O
Let X = (a, b), Y =(c, d) and Z = (e, f), ∀ a, c, e ∈ R0 and b, d, f ∈ R
X O (Y O Z) = (a, b) O(ce, de + f)
= (ace, bce + de + f)
(X O Y) O Z = (ac, bc+ d) O (e, f)
= (ace, (bc + d) e +f)
= (ace, bce + de + f)
Therefore, X O (Y O Z)= (X O Y) O Z, ∀ X, Y, Z ∈ A
(ii) Let E = (x, y) bethe identity element in A with respect to O, ∀ x ∈ R0 and y ∈ R
Such that,
X O E = X = E O X, ∀ X ∈ A
X O E = X and EOX = X
(ax, bx +y) = (a, b)and (xa, ya + b) = (a, b)
Considering (ax, bx +y) = (a, b)
ax = a
x = 1
And bx + y = b
y = 0 [since x = 1]
Considering (xa, ya +b) = (a, b)
xa = a
x = 1
And ya + b = b
y = 0 [since x = 1]
Therefore (1, 0) isthe identity element in A with respect to O.
(iii) Let F = (m, n) bethe inverse in A ∀ m ∈ R0 and n ∈ R
X O F = E and F O X =E
(am, bm + n) = (1, 0)and (ma, na + b) = (1, 0)
Considering (am, bm +n) = (1, 0)
am = 1
m = 1/a
And bm + n = 0
n = -b/a [since m =1/a]
Considering (ma, na +b) = (1, 0)
ma = 1
m = 1/a
And na + b = 0
n = -b/a
Therefore the inverseof (a, b) ∈ A with respect to Ois (1/a, -b/a)