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Hint- Use the simple property of product of two sets by the use of basic definition. The product of two sets contains every element of one set related to each and every element of another set.

Complete step-by-step solution -

Given that: $A = \left\{ {1,2,3} \right\}$ and $B = \left\{ {2,4} \right\}$

We have to find out: $A \times B,B \times A,A \times A,B \times B{\text{ and }}\left( {A \times B} \right) \cap \left( {B \times A} \right)$

As we know that for two general sets $X = \left\{ {a,b} \right\}{\text{ and }}Y = \left\{ {p,q} \right\}$ product of the set is:

$X \times Y = \left\{ {\left( {a,p} \right),\left( {a,q} \right),\left( {b,p} \right),\left( {b,q} \right)} \right\}$

So using the above general result proceeding for the given problem we have:

$

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

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

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

B \times B = \left\{ {2,4} \right\} \times \left\{ {2,4} \right\} = \left\{ {\left( {2,2} \right),\left( {2,4} \right),\left( {4,2} \right),\left( {4,4} \right)} \right\} \\

$

Now for $\left( {A \times B} \right) \cap \left( {B \times A} \right)$ that is the intersection of two sets, we have already found out $\left( {A \times B} \right){\text{ and }}\left( {B \times A} \right)$ in the above problem just we need to find out the common term between them.

From visualization of $\left( {A \times B} \right){\text{ and }}\left( {B \times A} \right)$ , we have only one common element i.e. $\left( {2,2} \right)$

So,

$\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right)} \right\}$

Hence, all the values of the set have been found out.

Note- The Cartesian product of two sets A and B, denoted A Ã— B, is the set of all possible ordered pairs where the elements of A are first and the elements of B are second. The intersection of two sets A and B, denoted by $A \cap B$ , is the set containing all elements of A that also belong to B (or equivalently, all elements of B that also belong to A).

Complete step-by-step solution -

Given that: $A = \left\{ {1,2,3} \right\}$ and $B = \left\{ {2,4} \right\}$

We have to find out: $A \times B,B \times A,A \times A,B \times B{\text{ and }}\left( {A \times B} \right) \cap \left( {B \times A} \right)$

As we know that for two general sets $X = \left\{ {a,b} \right\}{\text{ and }}Y = \left\{ {p,q} \right\}$ product of the set is:

$X \times Y = \left\{ {\left( {a,p} \right),\left( {a,q} \right),\left( {b,p} \right),\left( {b,q} \right)} \right\}$

So using the above general result proceeding for the given problem we have:

$

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

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

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

B \times B = \left\{ {2,4} \right\} \times \left\{ {2,4} \right\} = \left\{ {\left( {2,2} \right),\left( {2,4} \right),\left( {4,2} \right),\left( {4,4} \right)} \right\} \\

$

Now for $\left( {A \times B} \right) \cap \left( {B \times A} \right)$ that is the intersection of two sets, we have already found out $\left( {A \times B} \right){\text{ and }}\left( {B \times A} \right)$ in the above problem just we need to find out the common term between them.

From visualization of $\left( {A \times B} \right){\text{ and }}\left( {B \times A} \right)$ , we have only one common element i.e. $\left( {2,2} \right)$

So,

$\left( {A \times B} \right) \cap \left( {B \times A} \right) = \left\{ {\left( {2,2} \right)} \right\}$

Hence, all the values of the set have been found out.

Note- The Cartesian product of two sets A and B, denoted A Ã— B, is the set of all possible ordered pairs where the elements of A are first and the elements of B are second. The intersection of two sets A and B, denoted by $A \cap B$ , is the set containing all elements of A that also belong to B (or equivalently, all elements of B that also belong to A).

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