Question

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

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

Answer 1

Answer: (B)

$\frac{1}{16}$

Given det $\left(AB{A}^T\right)=8$

${\left|A\right|}^2.\left|B\right|=8$ ....... $\left|A\right|={\left|A\right|}^T$

and $\frac{\left|A\right|}{\left|B\right|}=8$

( On substituting the value of $\left|A\right|$ in teh equations we get)

$\Rightarrow \left|A\right|=4$ and $\left|B\right|=\frac{1}{2}$

$det(B{A}^{-1}.B^T)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$