**Cartesian product of two sets :**

If A and B are two non-empty sets, then the set of all ordered pairs (a, b) such that "a" belongs to A and "b" belongs to B, is called the Cartesian product of A and B, to be denoted by A x B.

Thus,

** A x B = { (a, b) : a ∈ A and b ∈ B }**

**And the Cartesian product of B and A, to be denoted by B x A.**

**Thus, **

**B x A = { (b, a) : a ∈ A and b ∈ B }**

And also,

**A x B ≠ B x A **

But,

**n(A x B) = n(B x A) **

If A is null set or B is null set, we define that A x B is null set.

The figure given below clearly illustrates the Cartesian product of two sets.

**Example 1 :**

Let A = { 1, 2, 3 }, B = { 4, 5 }. Find A x B and B x A

**Solution : **

A x B = { (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) }

B x A = { (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3) }

**Example 2 :**

If A x B = { (3, 2), (3, 4), (5, 2), (5, 4) }, find the sets A and B.

**Solution : **

Clear, A is the set of all first co-ordinates of A x B, while B is the set of all second co-ordinates of elements of A x B.

Therefore A = { 3, 5 } and B = { 2, 4 }

**Example 3 :**

Let P = { 1, 3, 6 }, Q = { 3, 5 }

Then, the product set

P x Q = { (1, 3), (1, 5), (3, 3), (3, 5), (6, 3), (6, 5) }

Q x P = { (3, 1), (3, 3), (3, 6), (5, 1), (5, 3), (5, 6) }

Notice that n(P) = 3, n(Q) = 2, n(PxQ) = 6 and n(QxP) = 6

And also,

**n(PxQ) = n(P) x n(Q) **

**n(QxP) = n(P) x n(Q) **

In Cartesian product, the ordered pairs (3, 5) and (5, 3) are not equal.

So,

**P x Q ≠ Q x P **

But,

**n(P x Q) = n(Q x P)**

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