# Solve the following

Question:

Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0\end{array}\right]$, where $\alpha \in R .$ Suppose $Q=\left[q_{i j}\right]$ is a matrix satisfying

$P Q=k I_{3}^{[}$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:

As $\mathrm{PQ}=\mathrm{KI} \Rightarrow \mathrm{Q}=\mathrm{kP}^{-1} \mathrm{I}$

Now $\mathrm{Q}=\frac{k}{|P|}(\operatorname{adj} P) I$

$\Rightarrow \mathrm{Q}=\frac{k}{(20+12 \alpha)}\left[\begin{array}{lll}- & - & - \\ - & - & (-3 \alpha-4)\end{array}\right]\left[\begin{array}{lll}1 & 0 & 0 \\ - & - & -\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

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

$\Rightarrow \frac{k}{(20+12 \alpha)}(-3 \alpha-4)=\frac{-k}{8}$

$3 \alpha=-3 \Rightarrow \alpha=-1$

Also $|Q|=\frac{k^{3}|I|}{|P|}$

$\Rightarrow \frac{k^{2}}{2}=\frac{k^{3}}{(20+12 \alpha)}$

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

$\Rightarrow k=4$