We know that if n(A)=p and n(B)=q, then n(A×B)=n(A)×n(B)=pq∴n(A×A)=n(A)×n(A)It is given that n(A×A)=9∴n(A)×n(A)=9⇒n(A)=3The ordered pairs (−1,0) and (0,1) are two of the nine elements of A×ANow, A×A={(a,a):a∈A} Therefore −1,0 and 1 are elements of ASince n(A)=3, so set A={−1,0,1}The remaining elements of set A×A are (−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0) and (1,1)