Question:
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 $A \times A$.
Solution:
We know that if $n(A)=p$ and $n(B)=q$, then $n(A \times B)=p q$.
$\therefore n(A \times A)=n(A) \times n(A)$
It is given that $n(A \times A)=9$
$\therefore n(A) \times n(A)=9$
$\Rightarrow n(\mathrm{~A})=3$
The ordered pairs $(-1,0)$ and $(0,1)$ are two of the nine elements of $A \times A$.
We know that $A \times A=\{(a, a): a \in A\}$. Therefore, $-1,0$, and 1 are elements of $A$.
Since $n(A)=3$, it is clear that $A=\{-1,0,1\}$.
The remaining elements of set $A \times A$ are $(-1,-1),(-1,1),(0,-1),(0,0)$,
$(1,-1),(1,0)$, and $(1,1)$