The trace of A is the sum of the diagonal elements: 1 + 4 = 5.

The eigenvalues of A are the diagonal elements: 2 and 3.

The union A ∪ B includes all elements from both sets, which are {1, 2, 3, 4, 6, 8}.

Set A is a subset of set B because all elements of A are contained in B. Therefore, A ⊆ B is True.

A is a subset of B if all elements of A are in B. Here, all elements of A are in B, so A ⊆ B is true.

The cardinality of the Cartesian product A × B is given by |A| * |B|. Here, |A| = 3 and |B| = 4, so |A × B| = 3 * 4 = 12.

A ∪ B = {1, 2, 3, 4}, which has 4 elements. The number of subsets is 2^4 = 16.

A × B = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)} as it is the Cartesian product of A and B.

The number of elements in the Cartesian product A × B is the product of the number of elements in A and B. Here, |A| = 3 and |B| = 2, so |A × B| = 3 * 2 = 6.

The difference of sets A and B, A - B, contains elements that are in A but not in B. Here, A - B = {1, 2}.

The symmetric difference A Δ B includes elements in either A or B but not in both, which are {1, 2, 4, 5}.

First, A ∩ B = {3}. Then, A ∪ {3} = {1, 2, 3}.

The Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B, which is {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}.

The number of subsets of a set with n elements is 2^n. Here, n = 3, so 2^3 = 8.

The power set of a set with n elements has 2^n subsets. Here, n=3, so the power set has 2^3 = 8 subsets.

The Cartesian product A × B consists of all ordered pairs (a, b) where a ∈ A and b ∈ B. Thus, A × B = {(1, x), (2, x)} for all odd integers x.

The intersection A ∩ B consists of common letters, which is {T}.

The intersection A ∩ B includes letters that are common in both words. The only common letter is 'A', so A ∩ B = {A}.

The intersection A ∩ B consists of common multiples, which is {0}.

Set A = {1, 2, 3, 4} and set B = {3, 4, 5, ...}. The intersection A ∩ B = {3, 4}.

The intersection A ∩ B includes all prime numbers less than 10, which are {2, 3, 5, 7}.

Set A = {1, 2, 3, 4} and set B = {1, 2}. Thus, A - B = {3, 4}.

The natural numbers less than 5 are {1, 2, 3, 4} and the prime numbers are {2, 3}. Their intersection is {2, 3}.

The intersection A ∩ B includes elements that are in both A and B. Here, A = {1, 2, 3, 4} and B = {1, 3}, so A ∩ B = {1, 3}.

Both sets A and B contain the same elements: {2, 3, 5, 7}. Therefore, A = B is True.

The set A consists of all prime numbers less than 10, which are {2, 3, 5, 7}.

The intersection A ∩ B is empty because no letter can be both a vowel and a consonant.

The union of sets A and B, A ∪ B, includes all even integers and all multiples of 3. Thus, A ∪ B = {0, 2, 3, 4, 6, 9, ...}.

A is the set of all even integers, while B is the set of multiples of 4. Not all even integers are multiples of 4 (e.g., 2 is even but not a multiple of 4), hence A is not a subset of B.

The only even prime number is 2, so A ∩ B = {2}.