The Cartesian product A times A has 9 element | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We know that if $n(A) = p,$ then $n(A 	imes A) = p^2.$Given $n(A 	imes A) = 9,$ so $p^2 = 9,$ which implies $n(A) = 3.$The elements of $A 	imes

Vedclass · Accepted Answer

$(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1)$

Vedclass · Answer

(A) We know that if $n(A) = p,$ then $n(A 	imes A) = p^2.$Given $n(A 	imes A) = 9,$ so $p^2 = 9,$ which implies $n(A) = 3.$The elements of $A 	imes A$ are of the form $(a, b)$ where $a, b \in A.$Since $(-1, 0) \in A 	imes A,$ we have $-1 \in A$ and $0 \in A.$Since $(0, 1) \in A 	imes A,$ we have $0 \in A$ and $1 \in A.$Thus,the set $A = \{-1, 0, 1\}.$The set $A 	imes A$ consists of $3 	imes 3 = 9$ elements:$A 	imes A = \{(-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1)\}.$The given elements are $(-1, 0)$ and $(0, 1).$The remaining elements are $(-1, -1), (-1, 1), (0, -1), (0, 0), (1, -1), (1, 0), (1, 1).$

The Cartesian product $A \times A$ has $9$ elements,among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.

Similar Questions

If $A = \{0, 1\}$ and $B = \{1, 0\}$,then $A \times B$ is equal to

If $A = \{ 2, 4, 5 \}$ and $B = \{ 7, 8, 9 \}$,then $n(A \times B)$ is equal to

If set $A$ has $p$ elements and set $B$ has $q$ elements,then the number of elements in $A \times B$ is:

Let $A = \{1, 2\}$ and $B = \{3, 4\}$. Find $A \times B$. How many subsets will $A \times B$ have? List them.

If $A = \{1, 2, 4\}, B = \{2, 4, 5\}, C = \{2, 5\},$ then $(A - B) \times (B - C)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module