JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Matrices

From $PQ=k{I}_3$ the matrix $Q$ equals $k{P}^{-1}=\frac{k}{|P|}\mathrm{adj}P$, so its $(2,3)$ entry is $\frac{k}{|P|}$ times the $(3,2)$ cofactor of $P$. That cofactor is $-\left|\begin{array}{cc}3& -2\\ 2& \alpha \end{array}\right|=-(3\alpha +4)$, and the determinant of $P$ is $12\alpha +20$. Setting the entry equal to $-\frac{k}{8}$ gives $\frac{3\alpha +4}{12\alpha +20}=\frac{1}{8}$, whence $12\alpha +20=8(3\alpha +4)$ and $\alpha =-1$, so $|P|=8$. The determinant of $Q=k{P}^{-1}$ is $\frac{k^3}{|P|}=\frac{k^3}{8}$; equating this to the given $\frac{k^2}{2}$ yields $k=4$. Hence ${\alpha }^2+k^2=1+16=17$.