Rbse Solutions for Class 11 maths Chapter 2 Exercise 2.1 | Cartesian Product of Sets

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 2 Exercise 2.1 . Master the concept of the Cartesian Product ($A \times B$) and $n(A \times B)$. Includes solving for unknown variables in ordered pairs, finding the number of subsets, and verifying properties like $A \times (B \cap C) = (A \times B) \cap (A \times C)$. Essential for building the foundation for Relations and Functions.

This exercise focuses on the Cartesian Product of Sets, denoted $A \times B$, which is the set of all possible ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

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1. Finding the values of $x$ and $y$

Given the equality of ordered pairs:

$$\left(\frac{x}{3} + 1, y – \frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)$$

For two ordered pairs to be equal, their corresponding components must be equal.

1. Equate the first components ($x$-coordinates):$$\frac{x}{3} + 1 = \frac{5}{3}$$$$\frac{x}{3} = \frac{5}{3} – 1$$$$\frac{x}{3} = \frac{5 – 3}{3} = \frac{2}{3}$$$$x = \mathbf{2}$$
2. Equate the second components ($y$-coordinates):$$y – \frac{2}{3} = \frac{1}{3}$$$$y = \frac{1}{3} + \frac{2}{3}$$$$y = \frac{1 + 2}{3} = \frac{3}{3}$$$$y = \mathbf{1}$$

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2. Finding the number of elements in $A \times B$

Given: $\mathbf{n(A) = 3}$ and $B = \{3, 4, 5\}$, so $\mathbf{n(B) = 3}$.

The number of elements in the Cartesian product $A \times B$ is given by the formula:

$$n(A \times B) = n(A) \cdot n(B)$$

$$n(A \times B) = 3 \times 3 = \mathbf{9}$$

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3. Finding $G \times H$ and $H \times G$

Given: $G = \{7, 8\}$ and $H = \{5, 4, 2\}$.

1. $G \times H$: Set of ordered pairs $(g, h)$ where $g \in G$ and $h \in H$.$$G \times H = \{(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)\}$$
2. $H \times G$: Set of ordered pairs $(h, g)$ where $h \in H$ and $g \in G$.$$H \times G = \{(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)\}$$(Note: $G \times H \ne H \times G$ since the ordered pairs are different.)

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4. True or False Statements

(i) If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m, n), (n, m)\}$.False. $P = Q = \{m, n\}$. The Cartesian product $P \times Q$ must contain $n(P) \cdot n(Q) = 2 \times 2 = 4$ elements.

$$P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$$

Correct Statement: If $P = \{m, n\}$ and $Q = \{n, m\}$, then $P \times Q = \{(m, n), (m, m), (n, n), (n, m)\}$.

(ii) If $A$ and $B$ are non-empty sets, then $A \times B$ is a non-empty set of ordered pairs $(x, y)$ such that $x \in A$ and $y \in B$.True. This is the correct definition of the Cartesian product for non-empty sets.

(iii) If $A = \{1, 2\}$, $B = \{3, 4\}$, then $A \times (B \cap \phi) = \phi$.True.

The intersection of any set with the null set ($\phi$) is the null set: $B \cap \phi = \phi$.

The Cartesian product of any set with the null set is the null set: $A \times \phi = \mathbf{\phi}$.

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5. Finding $A \times A \times A$

Given: $A = \{-1, 1\}$.

The triple Cartesian product $A \times A \times A$ is the set of all ordered triples $(a, b, c)$ where $a, b, c \in A$. $n(A \times A \times A) = 2^3 = 8$ elements.

$$A \times A \times A = \{(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1),$$

$$(1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)\}$$

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6. Finding Sets $A$ and $B$

Given: $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$.

• The set $A$ is the set of all first components of the ordered pairs.$$A = \{a, b, a, b\} = \mathbf{\{a, b\}}$$
• The set $B$ is the set of all second components of the ordered pairs.$$B = \{x, y, x, y\} = \mathbf{\{x, y\}}$$

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7. Verification of Set Properties

Given: $A = \{1, 2\}$, $B = \{1, 2, 3, 4\}$, $C = \{5, 6\}$, $D = \{5, 6, 7, 8\}$.

(i) Verify $A \times (B \cap C) = (A \times B) \cap (A \times C)$

1. LHS: $A \times (B \cap C)$First find the intersection: $B \cap C = \{1, 2, 3, 4\} \cap \{5, 6\} = \phi$.$$A \times (B \cap C) = A \times \phi = \mathbf{\phi}$$
2. RHS: $(A \times B) \cap (A \times C)$$$A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\}$$$$A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$$The intersection $(A \times B) \cap (A \times C)$ contains pairs that are common to both sets. Since the second components of $A \times B$ are $\{1, 2, 3, 4\}$ and the second components of $A \times C$ are $\{5, 6\}$, there are no common pairs.$$(A \times B) \cap (A \times C) = \mathbf{\phi}$$

Since LHS = RHS = $\phi$, the statement is Verified.

(ii) Verify $A \times C$ is a subset of $B \times D$

We must show that every element in $A \times C$ is also an element in $B \times D$.

1. Find $A \times C$:$$A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\}$$
2. Check elements against $B \times D$:For $A \times C \subset B \times D$, we need $a \in B$ for all $a \in A$ and $c \in D$ for all $c \in C$.
   • $A = \{1, 2\}$. Since $B = \{1, 2, 3, 4\}$, $\mathbf{A \subset B}$ is true.
   • $C = \{5, 6\}$. Since $D = \{5, 6, 7, 8\}$, $\mathbf{C \subset D}$ is true.

Since $A \subset B$ and $C \subset D$, it follows that $A \times C \subset B \times D$.

Checking the pairs:

• $(1, 5)$: $1 \in B$ and $5 \in D$. ($\checkmark$)
• $(1, 6)$: $1 \in B$ and $6 \in D$. ($\checkmark$)
• $(2, 5)$: $2 \in B$ and $5 \in D$. ($\checkmark$)
• $(2, 6)$: $2 \in B$ and $6 \in D$. ($\checkmark$)

The statement is Verified.

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8. Subsets of $A \times B$

Given: $A = \{1, 2\}$ and $B = \{3, 4\}$.

1. Write $A \times B$:$$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}$$
2. Number of Subsets:The number of elements in $A \times B$ is $n(A \times B) = 4$.The number of subsets of a set with $n$ elements is $2^n$.Number of subsets $= 2^4 = \mathbf{16}$.
3. List the Subsets:Let $S = A \times B$. The 16 subsets are:
   • 0 elements: $\phi$ (1)
   • 1 element: $\{(1, 3)\}, \{(1, 4)\}, \{(2, 3)\}, \{(2, 4)\}$ (4)
   • 2 elements: $\{(1, 3), (1, 4)\}, \{(1, 3), (2, 3)\}, \{(1, 3), (2, 4)\}, \{(1, 4), (2, 3)\}, \{(1, 4), (2, 4)\}, \{(2, 3), (2, 4)\}$ (6)
   • 3 elements: $\{(1, 3), (1, 4), (2, 3)\}, \{(1, 3), (1, 4), (2, 4)\}, \{(1, 3), (2, 3), (2, 4)\}, \{(1, 4), (2, 3), (2, 4)\}$ (4)
   • 4 elements: $\{(1, 3), (1, 4), (2, 3), (2, 4)\} = S$ (1)

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9. Finding Sets $A$ and $B$ from $A \times B$ elements

Given: $n(A) = 3$, $n(B) = 2$, and $(x, 1), (y, 2), (z, 1)$ are in $A \times B$, where $x, y, z$ are distinct.

• The set $A$ must be the set of all first components:$$A = \{x, y, z\}$$Since $x, y, z$ are distinct and $n(A)=3$, this is the complete set $A$.
• The set $B$ must be the set of all second components:$$B = \{1, 2, 1\} = \{1, 2\}$$Since $n(B)=2$, this is the complete set $B$.

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10. Finding Set $A$ and remaining elements of $A \times A$

Given: $n(A \times A) = 9$. This implies $n(A) \cdot n(A) = 9$, so $\mathbf{n(A) = 3}$.

Also, $(\mathbf{-1}, \mathbf{0})$ and $(\mathbf{0}, \mathbf{1})$ are elements of $A \times A$.

• Since the first and second components of $A \times A$ must come from $A$, the set $A$ must contain all the unique components found in the given ordered pairs:$$A \supset \{-1, 0, 1\}$$Since $n(A)=3$, the set $A$ must be exactly:$$A = \mathbf{\{-1, 0, 1\}}$$
• Remaining elements of $A \times A$:The total number of elements is 9. The elements are all possible pairs from $A \times A$.$$A \times A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}$$The elements found were $(-1, 0)$ and $(0, 1)$.The remaining 7 elements are:$$\{(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)\}$$