Question

Let A and B be two invertible matrices of order $3\times 3$. If det $\left({ABA}^T\right)=8$ and det $\left({AB}^{-1}\right)=8$, then det $\left({BA}^{-1}{B}^T\right)$ is equal to :

• A. $\frac{1}{4}$
• B. 16
• C. 1
• D. $\frac{1}{16}$

Answer 1

The correct option is D $\frac{1}{16}$

$|A{|}^2|B|=8$

$\frac{|A|}{|B|}=8$

Therefore,

$|A|=4$

$|B|=\frac{1}{2}$

Therefore,

$det\left(B{A}^{-1}{B}^T\right)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$