Question

Let \(A= \begin{bmatrix} 3&4\\ 1&-2 \end{bmatrix}\) and let \(AB= \begin{bmatrix} -5&41\\ 5&-13 \end{bmatrix}\). Then |B^T| =

• A. \(\frac{1}{14}\)
• B. 14
• C. 10
• D. -10
• E. -14

Answer 1

The Correct Option is B

We are given the matrix \( A \) and the matrix product \( AB \). We need to find \( |B^T| \), the determinant of the transpose of matrix \( B \).
Step 1: Use the property of determinants for matrix multiplication We know that: \[ |AB| = |A| \cdot |B| \] and \[ |B^T| = |B| \] Thus, to find \( |B^T| \), we first need to calculate \( |AB| \) and then \( |A| \).
Step 2: Calculate \( |A| \) and \( |AB| \) First, calculate the determinant of matrix \( A \): \[ A = \begin{bmatrix} 3 & 4 \\ 1 & -2 \end{bmatrix} \] \[ |A| = (3 \times -2) - (4 \times 1) = -6 - 4 = -10 \] Now, calculate the determinant of matrix \( AB \): \[ AB = \begin{bmatrix} -5 & 41 \\ 5 & -13 \end{bmatrix} \] \[ |AB| = (-5 \times -13) - (41 \times 5) = 65 - 205 = -140 \]
Step 3: Find \( |B| \) From the equation \( |AB| = |A| \cdot |B| \), we have: \[ |AB| = -140 \quad \text{and} \quad |A| = -10 \] Thus: \[ -140 = -10 \cdot |B| \] \[ |B| = \frac{-140}{-10} = 14 \]
Step 4: Conclusion Since \( |B^T| = |B| \), we have: \[ |B^T| = 14 \]

Answer 2

We are given \(A = \begin{bmatrix} 3 & 4 \\ 1 & -2 \end{bmatrix}\) and \(AB = \begin{bmatrix} -5 & 41 \\ 5 & -13 \end{bmatrix}\). We want to find |B^T|.

First, let's find the determinant of A:

|A| = (3)(-2) - (4)(1) = -6 - 4 = -10

Next, let's find the determinant of AB:

|AB| = (-5)(-13) - (41)(5) = 65 - 205 = -140

We know that |AB| = |A||B|. Therefore, |B| = \(\frac{|AB|}{|A|}\).

|B| = \(\frac{-140}{-10} = 14\)

Also, we know that |B^T| = |B|.

Therefore, |B^T| = 14.

Therefore, the answer is 14.