Mathematics - Determinant Question with Solution | TestHub

$$\begin{array}{rl}& \Rightarrow \alpha =\frac{64x}{3\times 16x}=\frac{4}{3}\\ & {\Delta }_1=\left|\begin{array}{ccc}1& \sin \alpha & \cos \alpha \\ 1& \cos \alpha & \sin \alpha \\ 1& -\sin \alpha & \cos \alpha \end{array}\right|\\ & =\left|\begin{array}{ccc}0& \sin \alpha -\cos \alpha & \cos \alpha -\sin \alpha \\ 0& \cos \alpha +\sin \alpha & \sin \alpha -\cos \alpha \\ 1& -\sin \alpha & \cos \alpha \end{array}\right|\\ & =(\sin \alpha -\cos \alpha {)}^2-\left({\cos }^2\alpha -{\sin }^2\alpha \right)\\ & ={\sin }^2\alpha +{\cos }^2\alpha -2\sin \alpha \cdot \cos \alpha -{\cos }^2\alpha \\ & =2{\sin }^2\alpha -2\sin \alpha \cdot \cos \alpha \\ & =2{\sin }^2\alpha (\sin \alpha -\cos \alpha )\end{array}$$ Now, $\sin \alpha -\cos \alpha =0$ for only $$\begin{array}{rl}\alpha & =\frac{\pi }{4}\text{ in }\left(0,\frac{\pi }{2}\right)\\ \therefore {\Delta }_1& =2(\sin \alpha )\times 0=0,\end{array}$$ since value of $\sin \alpha$ is finite for $\alpha \in \left(0,\frac{\pi }{2}\right)$ Hence non-trivivial solution for only one value of $\alpha$ in $\left(0,\frac{\pi }{2}\right)$ $$\begin{array}{rl}& \left|\begin{array}{ccc}\cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0\\ & \Rightarrow \left|\begin{array}{ccc}0& \sin \alpha & \cos \alpha \\ 0& \cos \alpha & \sin \alpha \\ 2\cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0\\ & \Rightarrow 2\cos \alpha \left({\sin }^2\alpha -{\cos }^2\alpha \right)=0\\ & \therefore \cos \alpha =0\text{ or }{\sin }^2\alpha -{\cos }^2\alpha =0\end{array}$$ But $\cos \alpha =0$ not possible for any value of $$\begin{array}{rl}& \alpha \in \left(0,\frac{\pi }{2}\right)\\ & \therefore {\sin }^2\alpha -{\cos }^2\alpha =0\Rightarrow \sin \alpha =-\cos \alpha ,\end{array}$$ which is also not possible for any value of $$\alpha \in \left(0,\frac{\pi }{2}\right)$$ Hence, there is no solution.