Question:
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(A B A^{T}\right)=8$ and $\operatorname{det}\left(A B^{-1}\right)=8$, then $\operatorname{det}\left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)$ is equal to :-
Correct Option: 2,
Solution:
$|\mathrm{A}|^{2} \cdot|\mathrm{B}|=8$ and $\frac{|\mathrm{A}|}{|\mathrm{B}|}=8 \Rightarrow|\mathrm{A}|=4$ and $|\mathrm{B}|=\frac{1}{2}$
$\therefore \operatorname{det}\left(\mathrm{BA}^{-1} \cdot \mathrm{B}^{\mathrm{T}}\right)=\frac{1}{4} \times \frac{1}{4}=\frac{1}{16}$
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