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

This question combines two basic determinant facts: the product rule and the transpose rule.

1. Option (A) $-16$: correct, found by dividing $\det[\mathbf{A}\cdot\mathbf{B}] = -80$ by $\det[\mathbf{A}] = 5$ to get $\det[\mathbf{B}] = -16$, and then using $\det[\mathbf{B}^T] = \det[\mathbf{B}]$.
2. Option (B) $-1/16$: wrong, this comes from mistakenly inverting the ratio instead of dividing $-80$ by $5$.
3. Option (C) $-400$: wrong, this comes from multiplying $-80$ and $5$ instead of dividing.
4. Option (D) $28$: wrong, this does not follow from any correct determinant rule applied to the given numbers.

Since $\det[\mathbf{A}\cdot\mathbf{B}] = \det[\mathbf{A}]\det[\mathbf{B}]$, we get $\det[\mathbf{B}] = -80/5 = -16$. A matrix's transpose never changes its determinant, so $\det[\mathbf{B}^T]$ stays $-16$, option (A).