Question

\( A \) and \( B \) are invertible matrices of the same order such that \( (AB)^{-1}= 8 \), if \( |A| = 2 \), then \( |B| \) is:

• A. 6
• B. 16
• C. 4
• D. \( \frac{1}{16} \)

Hint:

When working with determinants of products of matrices, recall that the determinant of the product is the product of the determinants, i.e., \( |AB| = |A| \cdot |B| \).

Answer 1

The Correct Option is D

We are given that \( A \) and \( B \) are invertible matrices, and we know the following: \[ % Option (AB)^{-1} = 8 \] From the properties of determinants, we know that: \[ |(AB)^{-1}| = \frac{1}{|AB|} \] We are also given \( |A| = 2 \). Using the property \( |AB| = |A| \cdot |B| \), we can write: \[ \frac{1}{|AB|} = 8 \] Thus: \[ |AB| = \frac{1}{8} \] Since \( |AB| = |A| \cdot |B| \), we substitute \( |A| = 2 \) into the equation: \[ |A| \cdot |B| = \frac{1}{8} \] Substituting \( |A| = 2 \): \[ 2 \cdot |B| = \frac{1}{8} \] Solving for \( |B| \): \[ |B| = \frac{1}{16} \] Therefore, the correct answer is option (D)