Question

Let $P$ be an $3\times 3$ matrix $P=\left[\begin{array}{ccc}3& -1& -2\\ 2& 0& \alpha \\ 3& -5& 0\end{array}\right]$ where $\alpha$ belongs to $\mathrm{ℝ}$. Suppose $Q=\left[q_{ij}\right]$ is a matrix $PQ=k{I}_3$for some non-zero $k\in \mathrm{ℝ}$. If $q_{23}=-k/8$ and $\left|Q\right|=k^2/2$, then ${\alpha }^2+k^2$ is equal to _________.

Answer 1

Explanation for the correct answer:

It is given that

$$\begin{gathered} PQ=k{I}_3 \\ \Rightarrow Q=k{P}^{-1}I \\ \Rightarrow Q=\frac{k}{\left|P\right|}(adjP)I \\ \Rightarrow Q=\frac{k}{20+12\alpha }\left[\begin{array}{ccc}\_& \_& \_\\ \_& \_& (-3\alpha -4)\\ \_& \_& \_\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right] \end{gathered}$$

But $q_{23}=\frac{-k}{8}$.

$$\begin{gathered} \Rightarrow \frac{k}{20+12\alpha }(-3\alpha -4)=\frac{-k}{8} \\ \Rightarrow \frac{1}{5+3\alpha }(-3\alpha -4)=\frac{-1}{2} \\ \Rightarrow -6\alpha -8=-5-3\alpha \\ \Rightarrow 3\alpha =-3 \\ \Rightarrow \alpha =-1 \end{gathered}$$

Also, $\left|Q\right|=\frac{k^3\left|I\right|}{\left|P\right|}$.

$$\begin{gathered} \Rightarrow \frac{k^2}{2}=\frac{k^3}{20+12\alpha } \\ \Rightarrow 2k=20+12(-1) \\ \Rightarrow k=\frac{8}{2} \\ \Rightarrow k=4 \end{gathered}$$

Hence,

$$\begin{gathered} {\alpha }^2+k^2={\left(-1\right)}^2+{\left(4\right)}^2 \\ =17 \end{gathered}$$

Hence, the correct answer is $17$.