Determinant of H = (3*4) - (2*1) = 12 - 2 = 10.

Det(H) = (5*3) - (4*2) = 15 - 8 = 7.

Det(I) = 1(1*1 - 2*0) - 2(0*1 - 2*1) + 1(0*0 - 1*1) = 1(1) - 2(-2) + 1(-1) = 1 + 4 - 1 = 4.

Det(I) = 2(0*3 - 2*1) - 3(1*3 - 2*0) + 1(1*1 - 0*0) = 2(0 - 2) - 3(3) + 1(1) = -4 - 9 + 1 = -12.

Det(J) = 1(1*1 - 2*0) - 2(0*1 - 1*1) + 1(0*0 - 1*1) = 1(1) - 2(-1) + 1(-1) = 1 + 2 - 1 = 2.

Det(J) = (5*8) - (6*7) = 40 - 42 = -2.

The determinant of F is 0 because the rows are linearly dependent.

The determinant of J is calculated as 1*(1*1 - 0*3) - 2*(0*1 - 0*2) + 1*(0*3 - 1*2) = 1 - 0 - 2 = -1.

Determinant = (1*4) - (2*3) = 4 - 6 = -2.

The determinant is calculated as \( 2*7 - 3*5 = 14 - 15 = -1 \).

The determinant is \( 5*8 - 6*7 = 40 - 42 = -2 \).

The determinant is calculated as \( 2(0*2 - 1*1) - 1(1*2 - 3*1) + 3(1*1 - 3*0) = 0 \).

The determinant is \( 2*4 - 1*3 = 8 - 3 = 5 \).

The determinant of an upper triangular matrix is the product of its diagonal elements: \( 1*1*1 = 1 \).

The determinant is calculated as \( 2*4 - 3*1 = 8 - 3 = 5 \).

The determinant is calculated as (2*7) - (3*5) = 14 - 15 = -1.

The rows are linearly dependent, hence the determinant is 0.

The determinant is \( 0*0 - 1*1 = 0 - 1 = -1 \).

The determinant of this matrix is 0 because the rows are linearly dependent.

The determinant is calculated as \( 3*4 - 2*1 = 12 - 2 = 10 \).

The determinant is 0 because the rows are linearly dependent.

The determinant is \( 2*7 - 3*5 = 14 - 15 = -1 \).

The determinant is calculated as \( ad - bc \).

The determinant of the identity matrix is 1.

The first determinant is -2 and the second is 0, so the total is -2 + 0 = -2.

The determinant evaluates to 0 as the rows are linearly dependent.

Using the determinant formula, we find that the determinant evaluates to 0.

Using the distance formula: d = √[(1 - 1)² + (5 - 2)²] = √[0 + 9] = √9 = 3.

Using the distance formula: d = √[(2 - 6)² + (3 - 8)²] = √[16 + 25] = √41.

Using the distance formula: d = √((6 - 6)² + (2 - 8)²) = √(0 + 36) = √36 = 6.