Mathematics - Matrices Question with Solution | TestHub

$PQ=kI$

$\left|P\right|.\left|Q\right|=k^3$

$\Rightarrow \left|P\right|=2k\ne 0\Rightarrow P$ is an invertible matrix

$\because PQ=kI$

$\therefore Q=k{P}^{-1}I$

$\therefore Q=\frac{ \mathrm{adj}.P}{2}$

$\because q_{23}=-\frac{k}{8}$

$\therefore \frac{-\left(3\alpha +4\right)}{2}=-\frac{k}{8}\Rightarrow k=4$

$\therefore \left|P\right|=2k\Rightarrow k=10+6\alpha \dots \left(\mathrm{i}\right)$

Put value of $k$ in $\left(\mathrm{i}\right)$.. we get $\alpha =-1$

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