Question

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

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

Answer 1

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

$\text{det}\left(AB{A}^T\right)=8\Rightarrow |A|\cdot |B|\cdot |A^T|=8\Rightarrow |A{|}^2|B|=8\cdots \left(1\right)[\because |A|=|A^T|]$

$\text{det}\left(A{B}^{-1}\right)=8$
$\Rightarrow |A|\cdot |B^{-1}|=8\Rightarrow \frac{|A|}{|B|}=8\Rightarrow |A|=8|B|\cdots \left(2\right)$

Put $|A|$ from equation $\left(2\right)$ in equation $\left(1\right)$, we get
$64|B{|}^2|B|=8\Rightarrow |B|=\frac{1}{2}\therefore |A|=8|B|=4$

Now, $\text{det}\left(B{A}^{-1}{B}^T\right)=|B|\cdot \frac{1}{|A|}\cdot |B^T|$
$=\frac{|B{|}^2}{|A|}=\frac{1}{16}$