If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
Practice Questions
1 question
Q1
If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
-24
24
0
12
Using the determinant formula, we find \( \det(B) = 1(1*0 - 4*6) - 2(0 - 4*5) + 3(0 - 1*5) = -24 \).
Questions & Step-by-step Solutions
1 item
Q
Q: If \( B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix} \), what is \( \det(B) \)?
Solution: Using the determinant formula, we find \( \det(B) = 1(1*0 - 4*6) - 2(0 - 4*5) + 3(0 - 1*5) = -24 \).
Steps: 10
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).
