Find the value of the determinant \( |D| \) where \( D = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \).

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Q: Find the value of the determinant \( |D| \) where \( D = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \).

Solution: The determinant is calculated as 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.

Steps: 10

Step 1: Write down the matrix D: D = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].
Step 2: Identify the elements of the first row: a = 1, b = 2, c = 3.
Step 3: Calculate the determinant using the formula: |D| = a * det(M1) - b * det(M2) + c * det(M3), where M1, M2, M3 are the 2x2 matrices formed by removing the row and column of each element.
Step 4: Calculate det(M1) for element a (1): M1 = [[1, 4], [6, 0]]. The determinant is (1*0 - 4*6) = 0 - 24 = -24.
Step 5: Calculate det(M2) for element b (2): M2 = [[0, 4], [5, 0]]. The determinant is (0*0 - 4*5) = 0 - 20 = -20.
Step 6: Calculate det(M3) for element c (3): M3 = [[0, 1], [5, 6]]. The determinant is (0*6 - 1*5) = 0 - 5 = -5.
Step 7: Substitute the determinants back into the formula: |D| = 1*(-24) - 2*(-20) + 3*(-5).
Step 8: Calculate each term: 1*(-24) = -24, -2*(-20) = +40, 3*(-5) = -15.
Step 9: Add the results together: -24 + 40 - 15 = 1.
Step 10: The final value of the determinant |D| is 1.