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 $(1,1)$