Q43 / 90 · JEE Main 2019 (11 Jan Shift 2) · Mathematics
Let $A$ and $B$ be two invertible matrices of order $3 \times 3$. If $\operatorname{det}\left(\mathrm{ABA}^{\mathrm{T}}\right)=8$ and det $\left(\mathrm{AB}^{-1}\right)=8$, then det $\left(\mathrm{BA}^{-1} \mathrm{~B}^{\mathrm{T}}\right)$ is
equal to
- A$\frac{1}{4}$
- B1
- Correct answerC$\frac{1}{16}$
- D16
Solution
Let $|A|=a,|B|=b$
$\left.\Rightarrow \quad \mid A^{\eta}\right\rceil=a\left|A^{-1}\right|=\frac{1}{a},\left|B^{\eta}\right|=b,\left|B^{-1}\right|=\frac{1}{b}$
$\because \quad A B A^{\eta}=8 \Rightarrow|A||B| \mid A^{\eta}-8 \ldots(1)$
$\Rightarrow \quad a \cdot b \cdot a=8 \Rightarrow a^{2} b=8$
$\because \quad A B^{-1}|=8 \Rightarrow H|\left|B^{-1}\right|=8 \Rightarrow a, \frac{1}{b}=8$
From (1) \& (2)
$a=4, b=\frac{1}{2}$
Then, $\left|B A^{-1} B^{\eta}\right|=|B|\left|A^{-1}\right|\left|B^{\eta}\right|=b \cdot \frac{1}{a} \cdot b=\frac{b^{2}}{a}=\frac{1}{16}$
Hence, the correct answer is option: 'C'.