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Q.The Cartesian product A×AA \times A has 9 elements among which are found (−1,0)(-1,0) and (0,1)(0,1). Find the set A and the remaining elements of A×AA \times A.

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Subjective· 2mImportance★★★★★
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A×AA\times A having 9 elements forces ∣A∣=3|A|=3; the given pairs pin down A={−1,0,1}A=\{-1,0,1\}, and the remaining 7 pairs follow by listing A×AA\times A in full.

Step 1: Find ∣A∣|A|.

If ∣A∣=n|A|=n, then ∣A×A∣=n2|A\times A| = n^2. We are told ∣A×A∣=9|A\times A|=9, so n2=9  ⟹  n=3n^2=9 \implies n=3.

Step 2: Find AA.

Every element of an ordered pair in A×AA\times A must itself belong to AA. From (−1,0)(-1,0) and (0,1)(0,1), we read off that −1, 0, 1∈A-1,\ 0,\ 1 \in A. That is already 3 distinct elements, matching ∣A∣=3|A|=3, so

A={−1,0,1}A = \{-1,0,1\}

Step 3: List all of A×AA\times A.

A×A={(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}A\times A = \{(-1,-1),(-1,0),(-1,1),(0,-1),(0,0),(0,1),(1,-1),(1,0),(1,1)\}

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