# EXERCISE 2.1(Class 11)

1. If , find the values of x and y.

Solution: Given,

As the ordered pairs are equal, the corresponding elements should also be equal.

Thus,

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving,

x + 3 = 5  and  3y – 2 = 1  [Taking L.C.M. and adding]

x = 2  and  3y = 3

Therefore, x = 2 and y = 1

2. If set A has 3 elements and set B = {3, 4, 5}, then find the number of elements in (A × B).

Solution: Given, set A has 3 elements, and the elements of set B are {3, 4, 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Therefore, the number of elements in (A × B) will be 9.

3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution: Given, G = {7, 8} and H = {5, 4, 2}

So,

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

4. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {mn} and Q = {nm}, then P × Q = {(mn), (nm)}

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (xy) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ

Solution:  (i) The statement is false.

The correct statement is

If P = {mn} and Q = {nm}, then

P × Q = {(mm), (mn), (n, m), (nn)}

(ii) True

(iii) True

5. If A = {–1, 1}, find A × A × A.

Solution: The A × A × A for a non-empty set A is given by

A × A × A = {(abc): ab∈ A}

Here, it is given A = {–1, 1}, A = {-1, 1}

A×A = {(-1, -1) (-1, 1), (1, -1), (1, 1)}

Again

A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

6. If A × B = {(ax), (ay), (bx), (by)}. Find A and B.

Solution: Given,

A × B = {(ax), (a, y), (bx), (by)}

Hence, A is the set of all first elements, and B is the set of all second elements.

Therefore, A = {ab} and B = {xy}

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

Solution: Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ

Thus,

L.H.S. = A × (B ∩ C) = A × Φ = Φ

And

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)}

Thus,

R.H.S. = (A × B) ∩ (A × C) = Φ

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) To verify: A × C is a subset of B × D

First,

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

And,

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

Hence verified

8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Solution: Given,

A = {1, 2} and B = {3, 4}

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

Number of elements in A × B is n(A × B) = 4

We know that,

If C is a set with n(C) = m, then n[P(C)] = 2m.

Thus, the set A × B has 24 = 16 subsets.

And these subsets are as given below:

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

9. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where xy and z are distinct elements.

Solution: Given,

n(A) = 3 and n(B) = 2;

and (x, 1), (y, 2), (z, 1) are in A × B.

We know that,

So, clearly, xy, and z are the elements of A;

and

1 and 2 are the elements of B.

As n(A) = 3 and n(B) = 2, it is clear that set A = {xyz} and set B = {1, 2}

10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

Solution: We know that,

If n(A) = and n(B) = q, then n(A × B) = pq.

Also, n(A × A) = n(A) × n(A)

Given,

n(A × A) = 9

So, n(A) × n(A) = 9

Thus, n(A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(a, a): a ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As n(A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)