Let A and B be two 3 times 3 non-singular mat | vedclass

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(A) Given, $\operatorname{det}(A^T B A) = 27$. Since $\operatorname{det}(A^T) = \operatorname{det}(A)$, we have $|A|^2 |B| = 27$  $(i)$. Also, $\opera

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$\frac{3}{32}$

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(A) Given, $\operatorname{det}(A^T B A) = 27$. Since $\operatorname{det}(A^T) = \operatorname{det}(A)$, we have $|A|^2 |B| = 27$  $(i)$. Also, $\operatorname{det}(A B^{-1}) = 8$, which implies $\frac{|A|}{|B|} = 8$, so $|B| = \frac{|A|}{8}$ $(ii)$. Substituting $(ii)$ into $(i)$: $|A|^2 \left(\frac{|A|}{8}\right) = 27 \Rightarrow |A|^3 = 216 \Rightarrow |A| = 6$. Then, $|B| = \frac{6}{8} = \frac{3}{4}$. We need to find $\operatorname{det}(B^T A^{-1} B) = |B^T| |A^{-1}| |B| = |B| \cdot \frac{1}{|A|} \cdot |B| = \frac{|B|^2}{|A|}$. Substituting the values: $\frac{(\frac{3}{4})^2}{6} = \frac{9/16}{6} = \frac{9}{96} = \frac{3}{32}$.

Let $A$ and $B$ be two $3 \times 3$ non-singular matrices such that $\operatorname{det}(A^T B A) = 27$ and $\operatorname{det}(A B^{-1}) = 8$. Then $\operatorname{det}(B^T A^{-1} B) = $

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