Class 11 Maths NCERT Solutions Chapter 2 Ex 2.1 – Cartesian Products of Sets

Since the ordered pairs are equal, corresponding elements are equal.

\dfrac{x}{3}+1=\dfrac{5}{3} \;\Rightarrow\; \dfrac{x}{3}=\dfrac{5}{3}-1=\dfrac{2}{3} \;\Rightarrow\; x=2

y-\dfrac{2}{3}=\dfrac{1}{3} \;\Rightarrow\; y=\dfrac{1}{3}+\dfrac{2}{3}=1

n(A) = 3 and n(B) = 3 (since B = {3, 4, 5} has 3 elements).

n(A × B) = n(A) × n(B) = 3 × 3 = 9

G × H pairs each element of G with each element of H, in that order.

H × G pairs each element of H with each element of G, in that order.

P = {m, n}, Q = {n, m}, so both P and Q equal {m, n}.

The full Cartesian product must pair every element of P with every element of Q: (m,m), (m,n), (n,m), (n,n).

This matches the definition of the Cartesian product exactly, for non-empty A and B.

B ∩ φ = φ, since nothing can be common with the empty set.

A × φ = φ, since the Cartesian product with the empty set is always empty.

A × A × A consists of every ordered triplet (a, b, c) with a, b, c ∈ {−1, 1}.

Since A has 2 elements, there are 2³ = 8 such triplets.

A = set of all first elements appearing in the ordered pairs = {a, b}.

B = set of all second elements appearing in the ordered pairs = {x, y}.

B ∩ C: B = {1,2,3,4}, C = {5,6} share no element, so B ∩ C = φ.

A × (B ∩ C) = A × φ = φ

A × B = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

A × C = {(1,5),(1,6),(2,5),(2,6)}

(A × B) ∩ (A × C): no pair is common to both lists, since the second coordinates never match.

(A × B) ∩ (A × C) = φ

A × C = {(1,5),(1,6),(2,5),(2,6)}

Since A ⊂ B (1, 2 ∈ {1,2,3,4}) and C ⊂ D (5, 6 ∈ {5,6,7,8}), every pair (a,c) with a ∈ A, c ∈ C also satisfies a ∈ B, c ∈ D.

A × B = {(1,3), (1,4), (2,3), (2,4)}

n(A × B) = 4, so the number of subsets = 2⁴ = 16.

All 16 subsets:

The second elements appearing in the given pairs are 1 and 2, and n(B) = 2, so these must be all of B.

B = {1, 2}

The first elements x, y, z are distinct, and n(A) = 3, so these must be all of A.

A = {x, y, z}

A × A has 9 elements, so n(A) × n(A) = 9, which gives n(A) = 3.

From (−1, 0) ∈ A × A: both −1 and 0 must be elements of A.

From (0, 1) ∈ A × A: both 0 and 1 must be elements of A.

Together, A must contain −1, 0 and 1 — exactly 3 elements, matching n(A) = 3.

All 9 elements of A × A: (−1,−1), (−1,0), (−1,1), (0,−1), (0,0), (0,1), (1,−1), (1,0), (1,1)

Removing the two given elements (−1,0) and (0,1) leaves the remaining 7.