Question

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 satisfying $PQ=k{I}_3$ for some non-zero $k\in R.$ If $q_{23}=-\frac{k}{8}$ and $|Q|=\frac{k^2}{2}$, then ${\alpha }^2+k^2$ is equal to

Answer 1

As $PQ=KI\Rightarrow Q=K{P}^{-1}I$
now
$Q=\frac{k}{|P|}\left(adj\right(P\left)\right)I\Rightarrow Q=\frac{k}{(20+12\alpha )}\left[\begin{array}{ccc}-& -& -\\ -& -& (-3\alpha -4)\\ -& -& -\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
$\because q_{23}=-\frac{k}{8}\Rightarrow \frac{k}{(20+12\alpha )}(-3\alpha -4)=-\frac{k}{8}\Rightarrow 2(3a+4)=5+3a$

$3\alpha =-3\Rightarrow \alpha =-1$
also $|Q|=\frac{k^3|I|}{|P|}\Rightarrow \frac{k^2}{2}=\frac{k^3}{(20+12\alpha )}$

$(20+12\alpha )=2k\Rightarrow 8=2k\Rightarrow k=4$

Hence ${\alpha }^2+k^2=17$