Cartesian product of sets - class-XI

$A=\left{a,b,c,d\right}$

$A\times B = {(1,3),(1,4),(2,3),(2,4)}$.

Ordered pair for co-ordinate of a point in $XY$ plane is written as $(x, y).$

An ordered pair is written in the form $(x-$coordinate, $y-$coordinate$)$.

 $(a, b) = (b, a)$ only if $a = b.$.

$y$ coordinate will be second entry in ordered pair.

Here, $x$ co-ordinate will be the first entry in ordered pair.

$a$ will be first entry in ordered pair $(a, b).$

If $x$ is first entry and $y$ is second, then its ordered pair will be $(x, y)$.

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Cartesian product of Set $A$ and $B$ is denoted by $A\times B$.

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

The cartesian product of two sets is the set of all possible ordered pairs whose first component is a member of the first set and whose second component is a member of the second set. 
Example:$ A = {1, 2} \quad B = {2}$
cartesian product, $A \times B = {(1,2), (2, 2)}$

To find B × A multiply each element of B with that of A & form an ordered pair.

$\left{ { 0,1 } \right} \times \left{ 1,0 \right} ={ \left{ (0,1),(0,0),(1,1),(1,0) \right}  }$

The given statement is false.

$A\cup B={1,2,3,8}$
$A\cap B={3}$
$\therefore (A\cup B)\times (A\cap B)$
$={(1,3),(2,3),(3,3),(8,3)}$

The elements will be
$(2,48),(48,2)$
$(4,46), (46,4)$
:
:
$(2n,50-2n), (50-2n,2n)$
Now we have
$2,4,6,8...$ upto $48$
This forms an A.P
The number terms is $24$.

$n(A\Delta B)=n(A)+n(B)-n(A\cap B)$

Here we don`t know the $n(A\cap B)$

So the answer cannot be Determined.

x has to be 3 and y has to be 5.

Given, $n\left( A \right) =4$ and $n\left( B \right) =5$ 

Product of two sets is the set of ordered pairs formed by mapping every
element from the first set to every element of the second set.

So, $ A \times B = $ { $ (1,2), (1,3), (1,4), (2,2), (2,3), (2,4) $ }

And $ B \times A = $ { $ (2,1), (2,2), (3,1), (3,2), (4,1), (4,2) $ }

Clearly, $ A \times B \neq B \times A $

Given $n(A)=3,$  and $n(B) =3$
Hence $n(A\times B) = 3\times 3=9$

If $A$ and $B$ are any two non-empty sets.
then $A\times B$$=\left{(x,y):x\in A  and  y\in B\right}$
As $A = \left{2,3\right}$ and $B = \left{1,2\right}$
$A \times B$$=\left{(2,1), (2,2), (3,1), (3,2)\right}$
Hence, option A.

$\displaystyle A=\left{ 2,4,5 \right} ,B=\left{ 7,8,9 \right} $

$n(A\times B)=n(A)\times n(B)$

As the ordered pairs are equal,
$ 3p + q = p - q $
$  => 2p = -2q $
$ => 2p + 2q = 0 $
$ => p + q = 0 $

$ n(P) $ can be of any value as we are not sure of $ n(Q) $
Hence, $ n(P) $ can take any of the given values.

If $A $ and $B$ are two non empty sets, then the Cartesian product $A \times B$ is set of all ordered pairs $(a,b)$ such that $a\in A$ and $b\in B$.

By equality of ordered pairs, we have
$(a - 2, b + 3) = (6, 8)$
On equating we get
$a - 2 = 6$
$a = 8$
$b + 3 = 8$
$b = 5$
So, the value of$ a = 8 $ and $ b = 5.$

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

In an ordered pair $(x,y)$, the first component is $x$ and the second component is $y$.

Given , $(x-y)=(-1,1)$

Given 

$A\times B = {(3,a),(3,-1),(3,0),(5,a),(5,-1),(5,0)}$

Given$(x,y)=(p,q)$
$(3x - 1, 9) = (11, p + 2)$
By equating 
$3x - 1 = 11$
$3x = 12$
$x = 4$
$9 = p + 2$
$p = 7$
So, the value of $x = 4, p = 7.$

Given $(x,y)=(p,q)$
$(4y + 5, 3p - 1) = (25, p + 1)$
On equating we get
$4y + 5 = 25$
$4y = 25 - 5$
$4y = 20$
$y = 5$
$3p - 1 = p + 1$
$3p - p = 1 + 1$
$2p = 2$
$p = 1$
So, the value of$ p = 1, y = 5$

$A= {2,3}$

$ A\times B = {(2,4),(2,a),(2,5),(1,4),(1,a),(1,5)}$

Say A has x elements and B has y elements in total. 

$A={0,1,2}, B={2,3}$ and $C={3,4}$
$A \cup B={0,1,2,3}$
No. of elements in $(A\cup B)\times C$ $=$ No. of elements in $(A \cup B)\times $ No. of elements in $C$$=4\times 2=8$

Given sets are $A={a,b,c,d}$ and ${x,y,z}$ 

If $n(A) =5$ and $n(B) = 4$

If set $A$ and set $B$ have $39$ common elements, then the number of common elements in set $A\times B$ and set $B\times A,=39^2$

Given $A={b,c,d}$ and  $B={x,y}$, then

For two sets, A and B the set difference of set B from set A is the set of all element in A but not in B.
Example:
$A={a,b,c,d}$
$B={c,d,e}$
$A-B={a,b}$

$A\times A \Rightarrow$ the first element will be from $A$ and the second element will also be from $A$.

If $n(A)=m$,and $n(B)=n$,then $n(A\times B)=mn$

C is not even logical while in the first 2 cases:

The cardinality of the Cartesian product of two finite sets A and B is the product of their individual cardinalities: n(A x B) = n(A) * n(B).

The Cartesian product distributes over intersection: (A intersect B) x C = (A x C) intersect (B x C).

Statement B 'A × B = B × A' is false when interpreted universally because Cartesian products are not commutative unless A = B. For A = {1, 2} and B = {3}, A × B = {(1,3), (2,3)} but B × A = {(3,1), (3,2)}, which are different sets. Options A, C, and D contain true statements (C has notation issues but the concept is correct).

The Cartesian product distributes over set difference: A x (B - C) = (A x B) - (A x C).

Given, $A={1,2}$ and $B={3,4}$.

as the given ordered pairs are equal,

If two ordered pairs are equal, then both x - coordinates are equal and both y-coordinates are equal.

Since  $ (a, 3)  =  (5,x)  $ then ,$ a = 5 , x = 6 $