What is Cartesian Product of Sets in Class 11 Maths?

Cartesian Product of Sets is part of Chapter 2 (Relations and Functions) in NCERT Class 11 Mathematics. This topic is fundamental for CBSE board exams and competitive tests.

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Cartesian Product of Sets

🎓 Class 11 Mathematics CBSE Theory Ch 2 — Relations and Functions ⏱ ~25 min

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2.1 Introduction

A large part of mathematics centres on recognising patterns -- observable links between quantities that change together. In everyday life, we encounter many such patterns that characterise relations^?: brother and sister, father and son, teacher and student. Similarly, in mathematics, we come across relations like "number \(m\) is less than number \(n\)," "line \(\ell\) is parallel to line \(m\)," or "set A is a subset of set B."

In all these situations, a relation pairs objects in a specific order. In this chapter, we shall learn how to link pairs of objects drawn from two sets and then study a special kind of relation called a function^?, which captures the idea of a precise, one-to-one correspondence between one quantity and another.

🧑‍🎓

Leibniz

G. W. Leibniz (1646 -- 1716)

Gottfried Wilhelm Leibniz, the renowned German mathematician and philosopher, was one of the first to use the word "function" in the analytical sense. His work laid foundations for the modern concept of functions in mathematics.

2.2 Cartesian Products of Sets

Suppose A is a set of 2 colours and B is a set of 3 objects:

\(A = \{\text{red, blue}\}\) and \(B = \{b, c, s\}\)

where \(b\), \(c\) and \(s\) represent a particular bag, coat and shirt, respectively. How many pairs of coloured objects can be formed from these two sets?

Proceeding systematically, we can see that there will be 6 distinct pairs:

(red, \(b\)), (red, \(c\)), (red, \(s\)), (blue, \(b\)), (blue, \(c\)), (blue, \(s\))

Fig 2.1 -- The 6 ordered pairs from \(A \times B\)

Recall that an ordered pair^? of elements taken from sets P and Q is written in the form \((p, q)\), where \(p \in P\) and \(q \in Q\). Two ordered pairs \((a, b)\) and \((c, d)\) are equal if and only if \(a = c\) and \(b = d\).

Definition 1 -- Cartesian Product

Given two non-empty sets P and Q, the Cartesian product \(P \times Q\) is the set of all ordered pairs of elements from P and Q: \[P \times Q = \{(p, q) : p \in P,\; q \in Q\}\] If either P or Q is the null set, then \(P \times Q\) will also be an empty set, i.e., \(P \times Q = \emptyset\).

From the illustration above, we note that:

\(A \times B = \{(\text{red}, b), (\text{red}, c), (\text{red}, s), (\text{blue}, b), (\text{blue}, c), (\text{blue}, s)\}\)

Now consider the two sets:

\(A = \{\text{DL, MP, KA}\}\), where DL, MP, KA represent Delhi, Madhya Pradesh and Karnataka respectively, and \(B = \{01, 02, 03\}\) representing codes for licence plates of vehicles issued by these states.

Fig 2.2 -- 9 pairs in the Cartesian product for licence plate codes

If the three states and Karnataka were making codes for licence plates that begin with an element from set A, the available pairs would be \(A \times B\) containing 9 such pairs, since there are 3 elements in each set and \(3 \times 3 = 9\) possible codes.

As a final illustration, consider the two sets \(A = \{a_1, a_2\}\) and \(B = \{b_1, b_2, b_3, b_4\}\):

\(A \times B = \{(a_1, b_1), (a_1, b_2), (a_1, b_3), (a_1, b_4), (a_2, b_1), (a_2, b_2), (a_2, b_3), (a_2, b_4)\}\)

Fig 2.3 -- The 8 ordered pairs formed from \(\{a_1, a_2\} \times \{b_1, b_2, b_3, b_4\}\)

The 8 ordered pairs thus formed can represent the position of points in the plane if A and B are subsets of the set of real numbers. The point at position \((a_i, b_j)\) will be distinct from the point at position \((b_j, a_i)\).

Key Remarks

(i) Two ordered pairs are equal if and only if the corresponding first elements are equal and the second elements are also equal.

(ii) If there are \(p\) elements in A and \(q\) elements in B, then there will be \(pq\) elements in \(A \times B\), i.e., if \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = pq\).

(iii) If A and B are non-empty sets and either A or B is an infinite set, then so is \(A \times B\).

(iv) \(A \times A \times A = \{(a, b, c) : a, b, c \in A\}\). Here \((a, b, c)\) is called an ordered triplet.

Worked Examples

Example 1

If \((x + 1, y - 2) = (3, 1)\), find the values of \(x\) and \(y\).

Solution

Since the ordered pairs are equal, the corresponding elements are equal.
Therefore \(x + 1 = 3\) and \(y - 2 = 1\).
Solving: \(x = 2\) and \(y = 3\).

Example 2

If \(P = \{a, b, c\}\) and \(Q = \{r\}\), form the sets \(P \times Q\) and \(Q \times P\). Are these two products equal?

Solution

By the definition of the Cartesian product:
\(P \times Q = \{(a, r), (b, r), (c, r)\}\)
\(Q \times P = \{(r, a), (r, b), (r, c)\}\)

Since, by the definition of equality of ordered pairs, the pair \((a, r)\) is not equal to the pair \((r, a)\), we conclude that \(P \times Q \neq Q \times P\).
However, the number of elements in each set will be the same.

Example 3

Let \(A = \{1, 2, 3\}\), \(B = \{3, 4\}\) and \(C = \{4, 5, 6\}\). Find:

(i) \(A \times (B \cap C)\) (ii) \((A \times B) \cap (A \times C)\) (iii) \(A \times (B \cup C)\) (iv) \((A \times B) \cup (A \times C)\)

Solution

(i) By the definition of the intersection, \(B \cap C = \{4\}\).
Therefore, \(A \times (B \cap C) = \{(1,4), (2,4), (3,4)\}\).

(ii) Now \(A \times B = \{(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)\}\)
and \(A \times C = \{(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}\).
Therefore, \((A \times B) \cap (A \times C) = \{(1,4), (2,4), (3,4)\}\).

(iii) Since \(B \cup C = \{3, 4, 5, 6\}\), we have
\(A \times (B \cup C) = \{(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)\}\).

(iv) Using the sets from part (ii) above, we obtain
\((A \times B) \cup (A \times C) = \{(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)\}\).

Note that (i) = (ii) and (iii) = (iv), illustrating the distributive laws of Cartesian products over intersection and union.

Example 4

If \(P = \{1, 2\}\), form the set \(P \times P \times P\).

Solution

\(P \times P \times P = \{(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)\}\).

Example 5

If \(\mathbf{R}\) is the set of all real numbers, what do the Cartesian products \(\mathbf{R} \times \mathbf{R}\) and \(\mathbf{R} \times \mathbf{R} \times \mathbf{R}\) represent?

Solution

The Cartesian product \(\mathbf{R} \times \mathbf{R}\) represents the set \(\{(x, y) : x, y \in \mathbf{R}\}\), which represents the coordinates of all the points in a two-dimensional plane.

The Cartesian product \(\mathbf{R} \times \mathbf{R} \times \mathbf{R}\) represents the set \(\{(x, y, z) : x, y, z \in \mathbf{R}\}\), which represents the coordinates of all the points in three-dimensional space.

Example 6

If \(A \times B = \{(p, q), (p, r), (m, q), (m, r)\}\), find A and B.

Solution

\(A\) = set of first elements = \(\{p, m\}\)
\(B\) = set of second elements = \(\{q, r\}\).

Interactive: Cartesian Product Builder

Enter elements of sets A and B (comma-separated) to see their Cartesian product

Set A:

Set B:

Click "Build" to see the Cartesian product.

Exercise 2.1

Q1. If \(\left(\frac{x}{3}+1, y-\frac{2}{3}\right) = \left(\frac{5}{3}, \frac{1}{3}\right)\), find the values of \(x\) and \(y\).

Since the ordered pairs are equal:
\(\frac{x}{3} + 1 = \frac{5}{3}\) ⇒ \(\frac{x}{3} = \frac{5}{3} - 1 = \frac{2}{3}\) ⇒ \(x = 2\).
\(y - \frac{2}{3} = \frac{1}{3}\) ⇒ \(y = \frac{1}{3} + \frac{2}{3} = 1\).
Therefore \(x = 2\) and \(y = 1\).

Q2. If the set A has 3 elements and the set B = \(\{3, 4, 5\}\), then find the number of elements in \(A \times B\).

Since \(n(A) = 3\) and \(n(B) = 3\),
\(n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9\).
The number of elements in \(A \times B\) is 9.

Q3. If \(G = \{7, 8\}\) and \(H = \{5, 4, 2\}\), find \(G \times H\) and \(H \times G\).

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

Q4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If \(P = \{m, n\}\) and \(Q = \{n, m\}\), then \(P \times Q = \{(m, n), (n, m)\}\).
(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 \(y \in B\).
(iii) If \(A = \{1, 2\}\), \(B = \{3, 4\}\), then \(A \times (B \cap \emptyset) = \emptyset\).

(i) False. \(P \times Q = \{(m,n), (m,m), (n,n), (n,m)\}\) (there are \(2 \times 2 = 4\) elements).

(ii) True.

(iii) True. Since \(B \cap \emptyset = \emptyset\), \(A \times \emptyset = \emptyset\).

Q5. If \(A = \{-1, 1\}\), find \(A \times A \times A\).

\(A \times A \times A = \{(a, b, c) : a, b, c \in \{-1, 1\}\}\)
\(= \{(-1,-1,-1), (-1,-1,1), (-1,1,-1), (-1,1,1), (1,-1,-1), (1,-1,1), (1,1,-1), (1,1,1)\}\).
There are \(2^3 = 8\) ordered triplets.

Q6. If \(A \times B = \{(a, x), (a, y), (b, x), (b, y)\}\), find \(A\) and \(B\).

\(A\) = set of all first elements = \(\{a, b\}\)
\(B\) = set of all second elements = \(\{x, y\}\).

Q7. Let \(A = \{1, 2\}\), \(B = \{1, 2, 3, 4\}\), \(C = \{5, 6\}\) and \(D = \{5, 6, 7, 8\}\). Verify that:
(i) \(A \times (B \cap C) = (A \times B) \cap (A \times C)\) (ii) \(A \times C\) is a subset of \(B \times D\).

(i) \(B \cap C = \emptyset\) (since B and C share no common elements).
So \(A \times (B \cap C) = A \times \emptyset = \emptyset\).
Also \(A \times B = \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}\)
and \(A \times C = \{(1,5),(1,6),(2,5),(2,6)\}\).
\((A \times B) \cap (A \times C) = \emptyset\) (no common pairs). Verified.

(ii) \(A \times C = \{(1,5),(1,6),(2,5),(2,6)\}\).
\(B \times D\) contains all pairs \((b, d)\) with \(b \in \{1,2,3,4\}\) and \(d \in \{5,6,7,8\}\).
Since \(\{1,2\} \subseteq \{1,2,3,4\}\) and \(\{5,6\} \subseteq \{5,6,7,8\}\), every element of \(A \times C\) is in \(B \times D\).
Therefore \(A \times C \subseteq B \times D\). Verified.

Q8. Let \(A = \{1, 2\}\) and \(B = \{3, 4\}\). Write \(A \times B\). How many subsets will \(A \times B\) have? List them.

\(A \times B = \{(1,3), (1,4), (2,3), (2,4)\}\).
\(n(A \times B) = 4\), so the number of subsets = \(2^4 = 16\).

The 16 subsets are: \(\emptyset\), \(\{(1,3)\}\), \(\{(1,4)\}\), \(\{(2,3)\}\), \(\{(2,4)\}\), \(\{(1,3),(1,4)\}\), \(\{(1,3),(2,3)\}\), \(\{(1,3),(2,4)\}\), \(\{(1,4),(2,3)\}\), \(\{(1,4),(2,4)\}\), \(\{(2,3),(2,4)\}\), \(\{(1,3),(1,4),(2,3)\}\), \(\{(1,3),(1,4),(2,4)\}\), \(\{(1,3),(2,3),(2,4)\}\), \(\{(1,4),(2,3),(2,4)\}\), \(\{(1,3),(1,4),(2,3),(2,4)\}\).

Q9. 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 \(A \times B\), find A and B, where \(x, y\) and \(z\) are distinct elements.

Since \((x,1), (y,2), (z,1) \in A \times B\), the first elements \(x, y, z\) must belong to A and the second elements \(1, 2\) must belong to B.
Since \(n(A) = 3\) and \(x, y, z\) are distinct: \(A = \{x, y, z\}\).
Since \(n(B) = 2\): \(B = \{1, 2\}\).

Q10. The Cartesian product \(A \times A\) has 9 elements among which are found \((-1, 0)\) and \((0, 1)\). Find the set A and the remaining elements of \(A \times A\).

Since \(n(A \times A) = 9 = n(A)^2\), we have \(n(A) = 3\).
Since \((-1, 0) \in A \times A\), both \(-1\) and \(0\) belong to A.
Since \((0, 1) \in A \times A\), both \(0\) and \(1\) belong to A.
Therefore \(A = \{-1, 0, 1\}\).

The remaining elements of \(A \times A\) are:
\(\{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1),(1,0),(1,1)\}\).

Activity: Exploring Cartesian Products

L3 Apply

Materials: Pencil, paper, coloured pens

Predict: If you have 4 shirts and 3 trousers, how many different outfits can you make? Can you list all possible pairs systematically?

Let S = {white, blue, red, green} be your set of shirts and T = {jeans, formal, shorts} be your set of trousers.
Draw a grid with shirts along one axis and trousers along the other.
Place a dot at every intersection. Each dot represents one outfit (an ordered pair).
Count the total dots. Verify that it equals \(n(S) \times n(T) = 4 \times 3 = 12\).
Now repeat with S x S. What do you notice about the number of elements?

Observe: The total number of outfits (ordered pairs) always equals the product of the number of elements in each set. This holds for any two finite sets. The grid visualization helps us see why the Cartesian product is named after Rene Descartes, who introduced the coordinate system!

Competency-Based Questions

Scenario: A college cafeteria offers a combo meal with one main course and one drink. The main course options are M = {Rice, Roti, Pasta} and drink options are D = {Tea, Coffee, Juice, Lassi}. The cafeteria manager wants to print all possible combo meal cards.

Q1. How many distinct combo meal cards can the manager print?

L1 Remember

(a) 7
(b) 12
(c) 16
(d) 3

Answer: (b) 12. \(n(M \times D) = n(M) \times n(D) = 3 \times 4 = 12\).

Q2. Is the combo (Tea, Rice) the same as (Rice, Tea) in the Cartesian product \(M \times D\)? Justify your answer using the concept of ordered pairs.

L2 Understand

Answer: No. In an ordered pair \((a, b)\), the first element comes from M and the second from D. (Rice, Tea) \(\in M \times D\), but (Tea, Rice) \(\notin M \times D\) since Tea \(\notin M\). The order matters in a Cartesian product.

Q3. The manager decides to add a dessert option: S = {Gulab Jamun, Ice Cream}. How many elements are in \(M \times D \times S\)? Analyse why this uses ordered triplets.

L4 Analyse

Answer: \(n(M \times D \times S) = 3 \times 4 \times 2 = 24\). Each element is an ordered triplet \((m, d, s)\) because we need to specify exactly three choices in a fixed order: first the main course, then the drink, then the dessert. Changing the order would change the meaning.

Q4. A student claims: "If \(A \times B = B \times A\), then A must equal B." Evaluate whether this claim is always true, sometimes true, or never true. Support your argument with examples.

L5 Evaluate

Answer: The claim is sometimes true.
- If \(A = B\), then certainly \(A \times B = B \times A\) (both are \(A \times A\)).
- If either A or B is empty, then \(A \times B = \emptyset = B \times A\) without requiring \(A = B\).
- If both are non-empty and unequal, say \(A = \{1\}\) and \(B = \{2\}\), then \(A \times B = \{(1,2)\}\) and \(B \times A = \{(2,1)\}\). These are not equal.
So the claim is true when \(A = B\) or when at least one set is empty.

Assertion--Reason Questions

Assertion (A): If \(A = \{1, 2\}\) and \(B = \{3, 4\}\), then the number of elements in \(A \times B\) is 4.
Reason (R): \(n(A \times B) = n(A) \times n(B)\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) -- Both A and R are true, and R explains A. Since \(n(A) = 2\) and \(n(B) = 2\), \(n(A \times B) = 2 \times 2 = 4\).

Assertion (A): \((3, 5) = (5, 3)\).
Reason (R): Two ordered pairs \((a, b)\) and \((c, d)\) are equal if and only if \(a = c\) and \(b = d\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) -- A is false because \(3 \neq 5\), so \((3,5) \neq (5,3)\). R is true and correctly states the condition for equality of ordered pairs.

Assertion (A): If \(A = \{1, 2\}\), then \(A \times A\) has 4 elements while \(A \times A \times A\) has 8 elements.
Reason (R): For a set with \(n\) elements, the Cartesian product with itself \(k\) times has \(n^k\) elements.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) -- Both true. \(n(A) = 2\), so \(n(A \times A) = 2^2 = 4\) and \(n(A \times A \times A) = 2^3 = 8\). R explains A.

Frequently Asked Questions

What is a Cartesian product of two sets?

The Cartesian product of sets A and B, denoted A x B, is the set of all ordered pairs (a, b) where a belongs to A and b belongs to B.

What is an ordered pair?

An ordered pair (a, b) is a pair of elements where the order matters. (a, b) is not equal to (b, a) unless a = b. Two ordered pairs are equal if and only if corresponding elements are equal.

How many elements does A x B have?

If set A has m elements and set B has n elements, then A x B has m times n elements.

What is the Cartesian product of a set with itself?

A x A consists of all ordered pairs (a, b) where both belong to A. If A has n elements, A x A has n-squared elements. For A = R, this represents the coordinate plane.

Is Cartesian product commutative?

No, the Cartesian product is generally not commutative. A x B differs from B x A unless A = B or one is empty.

Frequently Asked Questions — Relations and Functions

What is Cartesian Product of Sets in NCERT Class 11 Mathematics?

Cartesian Product of Sets is a key concept covered in NCERT Class 11 Mathematics, Chapter 2: Relations and Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Cartesian Product of Sets step by step?

To solve problems on Cartesian Product of Sets, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 2: Relations and Functions?

The essential formulas of Chapter 2 (Relations and Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Cartesian Product of Sets important for the Class 11 board exam?

Cartesian Product of Sets is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Cartesian Product of Sets?

Common mistakes in Cartesian Product of Sets include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Cartesian Product of Sets?

End-of-chapter NCERT exercises for Cartesian Product of Sets cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.

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