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Let $\mathrm{A}$ be a finite set of size $\mathrm{n}$. The number of elements in the power set of $A \times A$ is:

  1. $2^{n^2}$
  2. $\left(2^n\right)^2$
  3. $\left(2^2\right)^n$
  4. None of the above
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Detailed Video Solution 

Asking for number of elements in $A \times A$ is same as asking for number of relations possible over set $A.$

Every relation on set $A$ is subset of $A \times A.$ Every subset of $A \times A$ is a relation over set $A$.

Power set of $A \times A$ contains subsets of $A \times A$. So, Every element in power set of $A \times A$ is a relation on set $A.$

Hence, the cardinality of $P(A \times A)$ is same as number of relations possible on set $A.$

If $|A| = n, $ then $|P(A)| = 2^n, |A \times A| = n^2$, So, $|P(A \times A)| = 2^{n^2} = 2^{(n^2)}$

Detailed Video Solution 

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|A x A| = |A| * |A| = n*n = $n^{2}$

Let set X = A x A
So, |X| = |A x A| = $n^{2}$

Number of elements in the power set of X = Cardinality of Power Set of X

= $2^{\left | X \right |}$ = $2^{n^{2}}$

Correct Option: A

Note that $\left ( 2^{n} \right )^{2} = \left ( 2^{2} \right )^{n} = 2^{2n}$
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1 votes
The power set of a set \( A \) is the set of all subsets of \( A \), including the empty set and \( A \) itself. If \( A \) has \( n \) elements, then the power set of \( A \) will have \( 2^n \) elements.

Now, when we consider \( A \times A \), it means the Cartesian product of \( A \) with itself. If \( A \) has \( n \) elements, then \( A \times A \) will have \( n \times n = n^2 \) elements.

So, the number of elements in the power set of \( A \times A \) would be \( 2^{n^2} \).
Answer:

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