Let $P=\left[\begin{array}{ccc}3 & -1 & -2 \ 2 & 0 & \alpha \ 3 & -5 & 0\end{array} ight]$, where

Question

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

Professor PC Thomas Classes · Answer

(17) : Given $P Q=k I_{3} \Rightarrow|P Q|=k^{3}\left|I_{3}ight|$\ $\Rightarrow|P| \cdot|Q|=k^{3} \Rightarrow|P| \frac{k^{2}}{2}=k^{3}$\ $\Rightarrow|P|=2 k 
eq 0 \Rightarrow P$ is an invertible matrix.\ $	herefore \quad Q=k P^{-1} I=\frac{k \operatorname{adj} P}{|P|}=\frac{k}{2 k} \operatorname{adj} P$ $$\begin{equation*} \Rightarrow Q=\frac{\operatorname{adj} P}{2} 	ag{i} \end{equation*}$$ Now, $q_{23}=-\frac{k}{8} \Rightarrow \frac{-(3 \alpha+4)}{2}=-\frac{k}{8}$\ Also,\ $|P|=2 k \Rightarrow 12 \alpha+20=2 k \Rightarrow k=10+6 \alpha$\ $	herefore \quad$ From (i), we get\ $\frac{-(3 \alpha+4)}{2}=-\frac{(10+6 \alpha)}{8} \Rightarrow \alpha=-1$\ and $k=10+6(-1)=4$\ $	herefore \quad \alpha^{2}+k^{2}=(-1)^{2}+4^{2}=17$\