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 $A \times A$.
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 $A \times A$.
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 $n(A \times A)=9$
$\therefore n(A) \times n(A)=9$
$\Rightarrow n(A)=3$
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 element of set $A \times A$ are $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0),$ and $(1,1)$
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