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NCERT Exemplar · Q1

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

(i) A×BA \times B
(ii) B×AB \times A
(iii) B×BB \times B
(iv) A×AA \times A
Punjab PsebShort· 2mImportance★★★★★est
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✓ Free question

The Cartesian product A×BA \times B is the set of all ordered pairs (a,b)(a, b) where a∈Aa \in A and b∈Bb \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×BA \times B means "take every element from AA and pair it with every element from BB, in that order."

This matters because (a,b)≠(b,a)(a, b) \neq (b, a) unless a=ba = b. The first coordinate always comes from the first set, the second from the second set. If AA has mm elements and BB has nn elements, then A×BA \times B will have exactly m×nm \times n ordered pairs.

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

(i) Finding A×BA \times B

  1. Pair each element of AA with each element of BB.

    Start with −1∈A-1 \in A:

    • (−1,1)(-1, 1) and (−1,3)(-1, 3)

    Next, 2∈A2 \in A:

    • (2,1)(2, 1) and (2,3)(2, 3)

    Finally, 3∈A3 \in A:

    • (3,1)(3, 1) and (3,3)(3, 3)
  2. Collect all pairs.

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

We have 3×2=63 \times 2 = 6 ordered pairs, as expected.

(ii) Finding B×AB \times A

  1. Now reverse the role: pair each element of BB with each element of AA.

    Start with 1∈B1 \in B:

    • (1,−1)(1, -1), (1,2)(1, 2), (1,3)(1, 3)

    Next, 3∈B3 \in B:

    • (3,−1)(3, -1), (3,2)(3, 2), (3,3)(3, 3)
  2. Collect all pairs.

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

Again, 2×3=62 \times 3 = 6 ordered pairs.

Watch out

Notice that A×B≠B×AA \times B \neq B \times A in general. For instance, (−1,1)∈A×B(-1, 1) \in A \times B but (−1,1)∉B×A(-1, 1) \notin B \times A (since −1∉B-1 \notin B). The Cartesian product is not commutative.

(iii) Finding B×BB \times B

  1. Pair each element of BB with each element of BB itself.

    From 1∈B1 \in B:

    • (1,1)(1, 1) and (1,3)(1, 3)

    From 3∈B3 \in B:

    • (3,1)(3, 1) and (3,3)(3, 3)
  2. Collect all pairs.

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

We have 2×2=42 \times 2 = 4 ordered pairs.

(iv) Finding A×AA \times A

  1. Pair each element of AA with each element of AA.

    From −1∈A-1 \in A:

    • (−1,−1)(-1, -1), (−1,2)(-1, 2), (−1,3)(-1, 3)

    From 2∈A2 \in A:

    • (2,−1)(2, -1), (2,2)(2, 2), (2,3)(2, 3)

    From 3∈A3 \in A:

    • (3,−1)(3, -1), (3,2)(3, 2), (3,3)(3, 3)
  2. Collect all pairs.

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

We have 3×3=93 \times 3 = 9 ordered pairs.

Tip

When computing A×AA \times A or B×BB \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

The Cartesian products are:

  1. A×B={(−1,1),(−1,3),(2,1),(2,3),(3,1),(3,3)}A \times B = \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}
  2. B×A={(1,−1),(1,2),(1,3),(3,−1),(3,2),(3,3)}B \times A = \{(1, -1), (1, 2), (1, 3), (3, -1), (3, 2), (3, 3)\}
  3. B×B={(1,1),(1,3),(3,1),(3,3)}B \times B = \{(1, 1), (1, 3), (3, 1), (3, 3)\}
  4. A×A={(−1,−1),(−1,2),(−1,3),(2,−1),(2,2),(2,3),(3,−1),(3,2),(3,3)}A \times A = \{(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)\}

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