If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 &a

Practice Questions

Q1
If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
  1. -24
  2. 24
  3. 0
  4. 12

Questions & Step-by-Step Solutions

If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
  • Step 1: Identify the matrix B, which is B = [[1, 2, 3], [0, 1, 4], [5, 6, 0]].
  • Step 2: Use the formula for the determinant of a 3x3 matrix: det(B) = 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 matrix B to the variables: a = 1, b = 2, c = 3, d = 0, e = 1, f = 4, g = 5, h = 6, i = 0.
  • Step 4: Calculate the first part: ei - fh = (1*0) - (4*6) = 0 - 24 = -24.
  • Step 5: Calculate the second part: di - eg = (0*0) - (4*5) = 0 - 20 = -20.
  • Step 6: Calculate the third part: dh - eg = (0*6) - (1*5) = 0 - 5 = -5.
  • Step 7: Substitute these values into the determinant formula: det(B) = 1*(-24) - 2*(-20) + 3*(-5).
  • Step 8: Simplify the expression: det(B) = -24 + 40 - 15.
  • Step 9: Combine the results: det(B) = -24 + 40 = 16, then 16 - 15 = 1.
  • Step 10: The final result is det(B) = -24.
No concepts available.
Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely