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 $A\times A$

Answer 1

We know that if $n\left(A\right)=p$ and $n\left(B\right)=q$, then n($A\times B)=n(A)\times n(B)=pq$
$\therefore n(A\times A)=n\left(A\right)\times n\left(A\right)$
It is given that $n(A\times A)=9$
$\therefore n\left(A\right)\times n\left(A\right)=9$
$\Rightarrow n\left(A\right)=3$
The ordered pairs $(-1,0)$ and $(0,1)$ are two of the nine elements of $A\times A$
Now, $A\times A=\left\{\right(a,a):a\in A\}$
Therefore $-1,0$ and $1$ are elements of $A$
Since $n\left(A\right)=3$, so set $A=\{-1,0,1\}$
The remaining elements of set $A\times A$ are $(-1,-1),(-1,1),(0,-1),(0,0),(1,-1),(1,0)$ and $(1,1)$