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 $\left|A B^{- 1} a d j \left(A^{T} B\right)^{- 1}\right|=K,$ then the value of $16K$ is equal to

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, $\left|A B^{- 1} a d j \left(A^{T} B\right)^{- 1}\right|$
$\left|\right.A\left|\right.\times \frac{1}{\left|\right. B \left|\right.}\times \frac{1}{\left|\right. a d j \left(\right. A^{T} B \left.\right) \left|\right.}$
$\left|A\right|\times \frac{1}{\left|B\right|}\times \frac{1}{\left|A^{T} B\right|^{2}}=\frac{\left|A\right|}{\left|B\right|^{3} \left|A\right|^{2}}=\frac{1}{16}$
i.e. $16K=1$