Let $A = \{-1, 2, 3\}$ and $B = \{1, 3\}$. Determine (i) $A \times B$ (ii) $B \times A$ (iii) $B \times B$ (iv) $A \times A$

Let A = \-1, 2, 3\ and B = \1, 3\. Determine (i) A × B (ii) B × A (iii) B × B (iv) A × A

NCERT Exemplar · Q1

A = \{-1, 2, 3\}

(i)

A \times B

(ii)

B \times A

(iii)

B \times B

(iv)

A \times A

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Concept understanding — Cartesian Product

Cartesian Product: From Intuition to Definition

Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?

You can pair each size with each topping:

Small + Cheese, Small + Pepperoni, Small + Veggie
Medium + Cheese, Medium + Pepperoni, Medium + Veggie
Large + Cheese, Large + Pepperoni, Large + Veggie

That's $3 \times 3 = 9$ possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.

The Intuition

The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.

Think of it like a multiplication table for sets. If set $A$ has $m$ items and set $B$ has $n$ items, their Cartesian product has $m \times n$ items.

Note

The name comes from René Descartes, who used this idea to create the coordinate plane — every point $(x, y)$ on a graph is an element of the Cartesian product of the x-axis and y-axis.

The Precise Definition

Let $A$ and $B$ be two sets. The Cartesian product of $A$ and $B$ , written $A \times B$ , is the set of all ordered pairs $(a, b)$ where $a$ is from $A$ and $b$ is from $B$ .

A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}

The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."

Key Properties to Remember

Order matters: $A \times B$ is generally not the same as $B \times A$ . For example, if $A = \{1, 2\}$ and $B = \{x, y\}$ :
- $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$
- $B \times A = \{(x, 1), (x, 2), (y, 1), (y, 2)\}$
These are different sets because the pairs are ordered differently.
Size formula: If $|A| = m$ and $|B| = n$ , then $|A \times B| = m \times n$ . This holds even if one set is empty — then the product is empty.
Empty set: $A \times \emptyset = \emptyset$ and $\emptyset \times B = \emptyset$ . You can't form any pairs if one set has nothing to contribute.

Watch out

A common mistake: thinking $A \times B$ contains all possible combinations of elements from $A$ and $B$ without caring about order. But $(a, b)$ and $(b, a)$ are different pairs unless $a = b$ . Always treat ordered pairs as distinct based on position.

Examples to Cement the Idea

Example 1: $A = \{1, 2\}$ , $B = \{3, 4\}$

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

Four pairs, as expected ( $2 \times 2 = 4$ ).

Example 2: $A = \{a\}$ , $B = \{1, 2, 3\}$

A \times B = \{(a, 1), (a, 2), (a, 3)\}

Three pairs — every element of $B$ gets paired with the single element of $A$ .

Example 3: $A = \{0, 1\}$ , $B = \{0, 1\}$

A \times B = \{(0, 0), (0, 1), (1, 0), (1, 1)\}

This is the set of all possible 2-bit binary strings — a foundation for computer science.

Why This Matters

The Cartesian product is the mathematical backbone of:

Coordinate geometry: Every point $(x, y)$ in the plane is from $\mathbb{R} \times \mathbb{R}$ .
Database tables: A table's rows are elements of the Cartesian product of its column domains.
Probability: All possible outcomes of two independent events form a Cartesian product.
Functions: A function from $A$ to $B$ is a subset of $A \times B$ with special properties.

Important

The Cartesian product is not commutative ( $A \times B \neq B \times A$ in general), but it is associative: $(A \times B) \times C$ can be thought of as $A \times B \times C$ , the set of ordered triples. This extends naturally to any number of sets.

Quick Check for Yourself

If $A = \{1, 2\}$ and $B = \{2, 3\}$ , what is $A \times B$ ? What is $B \times A$ ? Are they the same?

Answer: $A \times B = \{(1, 2), (1, 3), (2, 2), (2, 3)\}$ ; $B \times A = \{(2, 1), (2, 2), (3, 1), (3, 2)\}$ . They share only $(2, 2)$ — the rest are different because the order of coordinates is swapped.

The Cartesian product of two sets is introduced at the very start of the NCERT Class 11 Mathematics chapter on Relations and Functions, and "Cartesian product of sets definition and examples" is a commonly searched foundational topic for CBSE board and JEE Main preparation. This concept also underlies coordinate geometry and the formal definition of a function, both of which are frequently tested in "relations and functions important questions".

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$ . We systematically pair each element of the first set with every element of the second set.

The Cartesian product captures a fundamental idea: how do we combine two sets to form ordered pairs? Unlike ordinary set operations (union, intersection), the Cartesian product creates a new kind of object — pairs where order matters. The notation $A \times B$ means "take every element from $A$ and pair it with every element from $B$ , in that order."

This matters because $(a, b) \neq (b, a)$ unless $a = b$ . The first coordinate always comes from the first set, the second from the second set. If $A$ has $m$ elements and $B$ has $n$ elements, then $A \times B$ will have exactly $m \times n$ ordered pairs.

Given $A = \{-1, 2, 3\}$ and $B = \{1, 3\}$ , let's work through each product.

(i) Finding $A \times B$

Pair each element of $A$ with each element of $B$ .

Start with $-1 \in A$ :
- $(-1, 1)$ and $(-1, 3)$
Next, $2 \in A$ :
- $(2, 1)$ and $(2, 3)$
Finally, $3 \in A$ :
- $(3, 1)$ and $(3, 3)$
Collect all pairs.

A \times B = \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}

We have $3 \times 2 = 6$ ordered pairs, as expected.

(ii) Finding $B \times A$

Now reverse the role: pair each element of $B$ with each element of $A$ .

Start with $1 \in B$ :
- $(1, -1)$ , $(1, 2)$ , $(1, 3)$
Next, $3 \in B$ :
- $(3, -1)$ , $(3, 2)$ , $(3, 3)$
Collect all pairs.

B \times A = \{(1, -1), (1, 2), (1, 3), (3, -1), (3, 2), (3, 3)\}

Again, $2 \times 3 = 6$ ordered pairs.

Watch out

Notice that $A \times B \neq B \times A$ in general. For instance, $(-1, 1) \in A \times B$ but $(-1, 1) \notin B \times A$ (since $-1 \notin B$ ). The Cartesian product is not commutative.

(iii) Finding $B \times B$

Pair each element of $B$ with each element of $B$ itself.

From $1 \in B$ :
- $(1, 1)$ and $(1, 3)$
From $3 \in B$ :
- $(3, 1)$ and $(3, 3)$
Collect all pairs.

B \times B = \{(1, 1), (1, 3), (3, 1), (3, 3)\}

We have $2 \times 2 = 4$ ordered pairs.

(iv) Finding $A \times A$

Pair each element of $A$ with each element of $A$ .

From $-1 \in A$ :
- $(-1, -1)$ , $(-1, 2)$ , $(-1, 3)$
From $2 \in A$ :
- $(2, -1)$ , $(2, 2)$ , $(2, 3)$
From $3 \in A$ :
- $(3, -1)$ , $(3, 2)$ , $(3, 3)$
Collect all pairs.

A \times A = \{(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)\}

We have $3 \times 3 = 9$ ordered pairs.

Tip

When computing $A \times A$ or $B \times B$ , you can organize the pairs in a grid: rows indexed by the first coordinate, columns by the second. This visual structure makes it easy to verify you haven't missed any pairs.

✓Final answer

A \times B = \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Set ANNUAL1 mark
$A \times B = \{(a, x), (a,y), (b,x), (b,y)\}$
›Reveal solutionSolution

$A=\{a,b\}$ , $B=\{x,y\}$ .

By definition, $A\times B$ consists of all ordered pairs $(p,q)$ with $p\in A$ and $q\in B$ . Given $A\times B=\{(a,x),(a,y),(b,x),(b,y)\}$ , the set of distinct first coordinates appearing is $\{a,b\}$ , so $A=\{a,b\}$ ; the set of distinct second coordinates is $\{x,y\}$ , so $B=\{x,y\}$ . Checking: $\{a,b\}\times\{x,y\}$ indeed produces exactly the four pairs $(a,x),(a,y),(b,x),(b,y)$ .

✓Final answer
$A=\{a,b\}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$n(A) = p$
›Reveal solutionSolution

A relation from A to B is a subset of $A\times B$ ; since $A\times B$ has $pq$ elements, it has $2^{pq}$ subsets, i.e. $2^{pq}$ possible relations.

Since $n(A)=p$ and $n(B)=q$ , the Cartesian product $A\times B$ contains exactly $n(A)\times n(B)=pq$ ordered pairs.

A relation from $A$ to $B$ is defined as any subset of $A\times B$ (including the empty relation and the full relation). A set with $pq$ elements has exactly $2^{pq}$ subsets.

✓Final answer
$2^{pq}$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2021Set ANNUAL1 mark
$A \times B = \{(a,1), (a,5), (1,2), (b,2), (b,5), (b,1)\}$
›Reveal solutionSolution

Reading the printed set as $A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\}$ (correcting a scan misprint of "(a,2)" as "(1,2)"), $A=\{a,b\}$ and $B=\{1,2,5\}$ , giving $B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}$ .

Noting a likely scan issue: the printed set $\{(a,1),(a,5),(1,2),(b,2),(b,5),(b,1)\}$ has one pair, $(1,2)$ , whose first coordinate breaks the pattern — in a genuine Cartesian product $A\times B$ every first coordinate must come from the same set $A$ , and here " $b$ " already appears paired with all three of $1,2,5$ while " $a$ " appears with only $1,5$ (missing $2$ ). This is consistent with "(1,2)" being a scanning misread of "(a,2)" (the two characters look similar in some fonts). With that one correction the six pairs read as a complete, consistent product:

$A\times B=\{(a,1),(a,2),(a,5),(b,1),(b,2),(b,5)\}.$

From this, $A=\{a,b\}$ (the first coordinates) and $B=\{1,2,5\}$ (the second coordinates), and indeed $n(A)\times n(B)=2\times3=6$ matches the 6 pairs given — confirming the correction.

To find $B\times A$ , pair every element of $B$ with every element of $A$ , in that order:

$B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}.$

✓Final answer
$B\times A=\{(1,a),(1,b),(2,a),(2,b),(5,a),(5,b)\}$

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Q.Let A={−1,2,3}A = \{-1, 2, 3\}A={−1,2,3} and B={1,3}B = \{1, 3\}B={1,3}. Determine

(i) Finding A×BA \times BA×B

(ii) Finding B×AB \times AB×A

(iii) Finding B×BB \times BB×B

(iv) Finding A×AA \times AA×A

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Q.Let $A = \{-1, 2, 3\}$ and $B = \{1, 3\}$ . Determine

(i) Finding $A \times B$

(ii) Finding $B \times A$

(iii) Finding $B \times B$

(iv) Finding $A \times A$

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