Question

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

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

Answer 1

Answer: (B)

$\frac{1}{16}$

Answer 2

$\left|A\right|^{2} .\left|B\right| =8$ and $\frac{\left|A\right|}{\left|B\right|} = 8 \Rightarrow\left|A\right| = 4$ and $\left|B\right| = \frac{1}{2}$
$\therefore \det\left(BA^{-1}.B^{T}\right) = \frac{1}{4} \times\frac{1}{4} = \frac{1}{16}$