let $P=\left[\begin{array}{ccc}3& -1& -2\\ 2& 0& \alpha \\ 3& -5& 0\end{array}\right]$ , where $\alpha \in R$ . Suppose $Q=\left[q_{ij}\right]$  is a matrix such that $PQ=KI$  where $K\in R$ , $K\ne 0$ and $I$ is the identity matrix of order 3. If  $q_{23}=-\frac{K}{8}$  and $\det \left(Q\right)=\frac{K^2}{2}$  then which of the following option (s) is/are correct?

PQ = KI  $\Rightarrow \frac{PQ}{K}=I\Rightarrow P^{-1}=\frac{Q}{K}$
Also $\left|P\right|=12\alpha +20$ 
Comparing the third elements of 2nd row on both sides, we get
$-\left(\frac{3\alpha +4}{12\alpha +20}\right)=\frac{1}{k}\times \frac{-K}{8}\Rightarrow 24\alpha +32=12\alpha +20\Rightarrow \alpha =-1$ 
Since  $\left|P\right|=8$
Also PQ = KI  $\Rightarrow \left|P\right|\left|Q\right|=K^3$ $\Rightarrow 8\times \frac{k^2}{2}=k^3\Rightarrow k=4\Rightarrow \left|Q\right|=\frac{k^2}{2}=8$
(B)  $4\alpha -K+8=4\times (-1)-4+8=0\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
(C) Now det (P adjQ) =  $\left|P\right|\left|adjQ\right|$
$=\left|P\right|{\left|Q\right|}^2=8\times 8^2=2^9$ 
(D) $\left|QadjP\right|=\left|Q\right|{\left|P\right|}^2=2^9$