Question

If $A$ is an invertible matrix of order $3$ and $B$ is another matrix of the same order as of $A,$ such that $\left|B\right|=2,A^T\left|A\right|B=A\left|B\right|B^T.$ If $\left|A{B}^{-1}adj{\left(A^{T}B\right)}^{-1}\right|=K,$ then the value of $16K$ is equal to

Answer 1

Answer: 1

$\because A^T\left|A\right|B=A\left|B\right|B^T$
Taking determinant on both sides, we get,
$\left|A^T\right|\left|A\right|\left|B\right|=\left|A\right|\left|B\right|\left|B^T\right|$
$\Rightarrow \left|A\right|=2$
Now, $\left|A{B}^{-1}adj{\left(A^{T}B\right)}^{-1}\right|$
$|A|\times \frac{1}{|B|}\times \frac{1}{|adj(A^{T}B)|}$
$\left|A\right|\times \frac{1}{|B|}\times \frac{1}{{|A^{T}B|}^2}=\frac{|A|}{{|B|}^3{|A|}^2}=\frac{1}{16}$
i.e. $16K=1$