Question

Two square matrices \( \mathbf{A} \) and \( \mathbf{B} \) of the same order are such that \( \det[\mathbf{A} \cdot \mathbf{B}] = -80 \) and \( \det[\mathbf{A}] = 5 \). Here, \( \mathbf{A} \cdot \mathbf{B} \) is the product of \( \mathbf{A} \) and \( \mathbf{B} \). The value of \( \det[\mathbf{B}^T] \) is

• A. -16
• B. \( -\dfrac{1}{16} \)
• C. -400
• D. 28

Hint:

Use det(AB) = det(A) det(B) first, then remember a matrix and its transpose share the same determinant.

Answer 1

The Correct Option is A

Step 1: Use the product rule for determinants.
For any two square matrices of the same order, \( \det[\mathbf{A} \cdot \mathbf{B}] = \det[\mathbf{A}] \cdot \det[\mathbf{B}] \).
Given \( \det[\mathbf{A} \cdot \mathbf{B}] = -80 \) and \( \det[\mathbf{A}] = 5 \).

Step 2: Solve for det[B].
\[ \det[\mathbf{B}] = \frac{\det[\mathbf{A}\cdot\mathbf{B}]}{\det[\mathbf{A}]} = \frac{-80}{5} = -16 \]
Step 3: Use the transpose rule.
A matrix and its transpose always have the same determinant, so \( \det[\mathbf{B}^T] = \det[\mathbf{B}] \).

Final Answer:
\( \det[\mathbf{B}^T] = -16 \). \[ \boxed{-16} \]