D is the mother of E, and since E is the mother of F, D is F's grandmother.

D is the sister of E, and since E is the brother of F, D is also the sister of F.

D is the sister of E, and since E is the brother of F, D is also F's sister.

F must be last since D must come before E and E must come before F.

E · F = 1*1 + 1*2 + 1*3 = 1 + 2 + 3 = 6.

E · F = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

E · F = 2*1 + (-1)*2 + 3*0 = 2 - 2 + 0 = 0.

E · F = 5*1 + 5*2 + 5*3 = 5 + 10 + 15 = 30.

E · F = x*2 + y*3 + z*4 = 10 gives the equation 2x + 3y + 4z = 10.

Matrix E is singular because its determinant is 0 (1*4 - 2*2 = 0).

The determinant of E can be calculated using the rule of Sarrus or cofactor expansion, resulting in 0.

The determinant of E is calculated as (a*d) - (b*c) = ad - bc.

If 1 is included, we can choose from the remaining 4 elements (2, 3, 4, 5). The number of subsets is 2^4 = 16.

The total number of subsets is 2^4 = 16. Proper subsets exclude the set itself, so there are 16 - 1 = 15 proper subsets.

The power set of a set with n elements has 2^n elements. Here, n = 2, so 2^2 = 4.

The subsets containing 1 are {1}, {1, 2}, and the empty set is not included.

The total number of subsets is 2^4 = 16. Proper subsets exclude the set itself, so 16 - 1 = 15.

If 'a' is included, we can choose from the remaining elements {b, c}. The subsets containing 'a' are {a}, {a, b}, {a, c}, and {a, b, c}, totaling 4 subsets.

The union of two sets includes all unique elements from both sets. Here, the union is {a, b, c, d}.

The power set of a set with n elements has 2^n subsets. For E, n = 2, so the size of the power set is 2^2 = 4.

The prime numbers less than 10 are 2, 3, 5, and 7. Therefore, E = {2, 3, 5, 7}.

The prime numbers less than 10 are {2, 3, 5, 7}. The power set has 2^4 = 16 elements.

F's share = 4 parts. Total parts = 3 + 4 = 7. Total profit = (7/4) * 28000 = $49,000.

Tom is Emily's brother.

Since every student in the class passed the exam and John is a student, we can conclude that John passed the exam.

The inverse of the identity matrix F is itself, so F^(-1) = F.

Det(F) = 1(1*0 - 4*6) - 2(0*0 - 4*5) + 3(0*6 - 1*5) = 1(0 - 24) - 2(0 - 20) + 3(0 - 5) = -24 + 40 - 15 = 1.

The determinant of F is calculated as (1*4) - (2*2) = 4 - 4 = 0.

The second row is a multiple of the first row, so there is only one linearly independent row. Therefore, the rank of F is 1.

The trace of F is the sum of the diagonal elements: 1 + 5 = 6.