HOW TO FIND 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 ∈ A, b ∈ B is called the Cartesian Product of A and B, and is denoted by A x B .

Thus, A x B = { (a,b) |a ∈ A,b ∈ B } 

A x B is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of A and the second coordinate is an element of B.

B x A is the set of all possible ordered pairs between the elements of A and B such that the first coordinate is an element of B and the second coordinate is an element of A.

If a = b, then (a, b)  =  (b, a).

The 'Cartesian Product' is also referred as 'Cross Product'.

In general

AxB  ≠  BxA,

But,

n(A x B)  =  n(B x A)

AxB = ∅, if and only if A = ∅ or B = ∅.

If n(A) = p and n(B) = q ,then

n(AxB)  =  pq

Practice Problems

Problem 1 :

Find AxB , AxA and BxA : 

A  =  {2, -2, 3} and B  =  {1, -4}

Solution :

A  =  {2, -2, 3} and B  =  {1, -4}.

AxB  =  {(2, 1), (2, -4), (-2, 1), (-2, -4), (3, 1), (3, -4)}

A  =  {2, -2, 3} and A  =  {2, -2, 3}.

AxA  =  {(2, 2), (2, -2), (2, 3), (-2, 2), (-2, -2), (-2, 3), 

(3, 2), (3, -2), (3, 3)}

B  =  {1, -4} and A  =  {2, -2, 3}. 

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

Problem 2 :

Find AxB , AxA and BxA : 

A  =  B  =  {p, q}

Solution :

A  =  {p, q} and B  =  {p, q}

AxB  =  {(p, p), (p, q), (q, p), (q, q)}

A  =  {p, q}, A  =  {p, q}

AxA  =  {(p, p) (p, q) (q, p) (q, q)}

B  =  {p, q} and B  =  {p, q}

BxA  =  {(p, p), (q, p), (p, q), (q, q)}

Problem 3 :

Find AxB , AxA and BxA : 

A  =  {m, n} and B  =  ∅

Solution :

Because B = ∅,

AxB  =  ∅

BxA  =  ∅

A  =  {m, n} and A  =  {m, n}

AxA  =  {(m, m), (m, n), (n, m), (n, n)}

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