# Solve this following

Question:

Let $\mathrm{P}=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in \mathrm{R}$. Suppose

$\mathrm{Q}=\left[\mathrm{q}_{\mathrm{ij}}\right]$ is a matrix satisfying $\mathrm{PQ}=\mathrm{kI}_{3}$ for some

non-zero $\mathrm{k} \in \mathrm{R} .$ If $\mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$ and $|\mathrm{Q}|=\frac{\mathrm{k}^{2}}{2}$,

then $\alpha^{2}+\mathrm{k}^{2}$ is equal to

Solution:

$\mathrm{PQ}=\mathrm{kI}$

$|\mathrm{P}| .|\mathrm{Q}|=\mathrm{k}^{3}$

$\Rightarrow|\mathrm{P}|=2 \mathrm{k} \neq 0 \Rightarrow \mathrm{P}$ is an invertible matrix

$\because \mathrm{PQ}=\mathrm{kI}$

$\therefore \mathrm{Q}=\mathrm{kP}^{-1} \mathrm{I}$

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

$\because \mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$

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

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

Put value of $k$ in (i).. we get $\alpha=-1$