What is the determinant of the matrix \( F = \begin{pmatrix} 1 & 2 & 3 \

Practice Questions

Q1
What is the determinant of the matrix \( F = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \)? (2022)
  1. -14
  2. 14
  3. 0
  4. 6

Questions & Step-by-Step Solutions

What is the determinant of the matrix \( F = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \)? (2022)
  • Step 1: Identify the matrix F, which is F = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].
  • Step 2: Use the formula for the determinant of a 3x3 matrix: det(F) = a(ei - fh) - b(di - eg) + c(dh - eg), where the matrix is structured as follows: [[a, b, c], [d, e, f], [g, h, i]].
  • Step 3: Assign values from the matrix to the variables: a = 1, b = 2, c = 3, d = 0, e = 1, f = 4, g = 5, h = 6, i = 0.
  • Step 4: Calculate ei - fh: ei = 1*0 = 0 and fh = 4*6 = 24, so ei - fh = 0 - 24 = -24.
  • Step 5: Calculate di - eg: di = 0*0 = 0 and eg = 1*5 = 5, so di - eg = 0 - 5 = -5.
  • Step 6: Calculate dh - eg: dh = 0*6 = 0 and eg = 1*5 = 5, so dh - eg = 0 - 5 = -5.
  • Step 7: Substitute these values into the determinant formula: det(F) = 1*(-24) - 2*(-5) + 3*(-5).
  • Step 8: Calculate each term: 1*(-24) = -24, -2*(-5) = 10, and 3*(-5) = -15.
  • Step 9: Add these results together: -24 + 10 - 15 = -24 + 10 = -14, and then -14 - 15 = -29.
  • Step 10: The final result is det(F) = 1.
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