Question
Question: If \(n\left( A\times B \right)\cap \left( C\times D \right)=n\left[ \left( A\cap C \right)\times \le...
If n(A×B)∩(C×D)=n[(A∩C)×(B∩D)]
A.Statement-1 is true, statement-2 is true. Statement-2 is a correct explanation for statement-1.
B.Statement-1 is true, statement-2 is true. Statement-2 is not a correct explanation for statement-1.
C.Statement-1 is true, statement-2 is false.
D.Statement-1 is true, statement-2 is true.
Solution
We will first assume the elements of the set A and set B. Then, we will find A×B and B×A and we will check the number of common elements in the set A×B and B×A. If we will get 92 elements common, then the first statement is true. We will then check statement 2 and then its relation with statement 1.
Complete step-by-step answer:
Let the set A be\left\\{ 1,2,3,4,5,6,7,8,9,15 \right\\} and set B be\left\\{ 1,2,3,4,5,6,7,8,9,25 \right\\}. Here set A and set B have 9 elements in common and both sets are of 10 elements.
We will first find the product of set A and set B i.e. A×B
A\times B=\left\\{ 1,2,3,4,5,6,7,8,9,15 \right\\}\times \left\\{ 1,2,3,4,5,6,7,8,9,25 \right\\}
Thus, the product is given by:-
A\times B=\left\\{ \begin{aligned}
& \left( 1,1 \right),.............,\left( 1,25 \right) \\\
& \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,25 \right) \\\
& \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,25 \right) \\\
& ............ \\\
& \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,25 \right) \\\
& \left( 15,1 \right),\left( 15,2 \right),..........,\left( 1,25 \right) \\\
\end{aligned} \right\\}
Similarly, we will find the product
B\times A=\left\\{ \begin{aligned}
& \left( 1,1 \right),.............,\left( 1,15 \right) \\\
& \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,15 \right) \\\
& \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,15 \right) \\\
& ............ \\\
& \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,15 \right) \\\
& \left( 25,1 \right),\left( 15,2 \right),..........,\left( 1,15 \right) \\\
\end{aligned} \right\\}
We have observed that there are 81 elements common in set and, these are:-
\left\\{ \begin{aligned}
& \left( 1,1 \right),.............,\left( 1,9 \right) \\\
& \left( 2,1 \right),\left( 2,2 \right),...........,\left( 1,9 \right) \\\
& \left( 3,1 \right),\left( 3,2 \right),..........,\left( 1,9 \right) \\\
& ............ \\\
& \left( 9,1 \right),\left( 9,2 \right),...........,\left( 1,9 \right) \\\
\end{aligned} \right\\}
Hence, the statement 1 is true.
Now, we will check statement 2.
The given statement is
n(A×B)∩(C×D)=n[(A∩C)×(B∩D)]
Yes, the statement is true. Let’s understand it with the same example.
The products of two sets are A×B and B×A.
We know, number of elements common in both sets A×B and B×A are92 which is given by n(A×B)∩(B×A)=92
We know, number of elements common in set A and set B is 9 which is given by A∩Band we know the product of their intersection will have 81 elements i.e. n[(A∩B)×(B∩A)]=92
Thus, we can write
n(A×B)∩(B×A)=n[(A∩B)×(B∩A)], the statement is similar to the statement 2 i.e.n(A×B)∩(C×D)=n[(A∩C)×(B∩D)]
Hence, statement 2 is also true.
Therefore, Statement-1 is true, statement-2 is true. Statement-2 is a correct explanation for statement-1.
Thus, the correct option is A.
Note: Here we have calculated the product of two sets and also found their intersection. Let set A and set B are two sets, then the Cartesian product of set A and set B is denoted by A×B and it is defined as a set which consists of all ordered pair (a,b) for which a∈A and b∈B and intersection of set A and set B is denoted by A∩B and it shows the common elements between the set A and set B.