Question

The cartesian product $A\times A$ has $9$ elements among which are found $(-1,0)$ and $(0,1),$ then set $A$ = ?

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

Answer 1

Answer: (B)

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

Answer 2

Given, $n(A\times A)=9\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,n(A)\times n(A)=9$ which is possible when, $n(A)=3.$ i.e., set A contain only three elements i.e., $A=\\{1,-1,0\\}$ So, $A\times A$ contain $(-1,0)$ and $(0,\,1)$