Question

Let $A$ be a non-empty set such that $A\times A$ has $9$ elements among which are found $(-1,0)$ and $(0,1)$, then

• A. $A=\{-1,0\}$
• B. $A=\{0,1\}$
• C. $A=\{-1,0,1\}$
• D. $A=\{-1,1\}$

Answer 1

Answer: (C)

$A=\{-1,0,1\}$

Given $A\times A$ has $9$ elements $\Rightarrow n\left(A\right)=3$
Elements of $A\times A$ are $(-1,0)$ & $(0,1)$.
$\therefore$ $A$ needs to contain $-1,0$ and $1$.
$\therefore$ Since $n\left(A\right)=3\Rightarrow A=\{-1,0,1\}$