Statement-1: The system of linear equations … has a non-trivial solution for only one value of α lying in the interval (0, π 2 ). Statement-2: The…

$$\begin{aligned} & \Rightarrow \alpha=\frac{64 x}{3 \times 16 x}=\frac{4}{3} \\ & \Delta_1=\left|\begin{array}{ccc} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & \cos \alpha \end{array}\right| \\ & =\left|\begin{array}{ccc} 0 & \sin \alpha-\cos \alpha & \cos \alpha-\sin \alpha \\ 0 & \cos \alpha+\sin \alpha & \sin \alpha-\cos \alpha \\ 1 & -\sin \alpha & \cos \alpha \end{array}\right| \\ & =(\sin \alpha-\cos \alpha)^2-\left(\cos ^2 \alpha-\sin ^2 \alpha\right) \\ & =\sin ^2 \alpha+\cos ^2 \alpha-2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha \\ & =2 \sin ^2 \alpha-2 \sin \alpha \cdot \cos \alpha \\ & =2 \sin ^2 \alpha(\sin \alpha-\cos \alpha) \end{aligned}$$​⇒α=3×16x64x​=34​Δ1​=​111​sinαcosα−sinα​cosαsinαcosα​​=​001​sinα−cosαcosα+sinα−sinα​cosα−sinαsinα−cosαcosα​​=(sinα−cosα)2−(cos2α−sin2α)=sin2α+cos2α−2sinα⋅cosα−cos2α=2sin2α−2sinα⋅cosα=2sin2α(sinα−cosα)​ Now, $\sin \alpha-\cos \alpha=0$sinα−cosα=0 for only $$\begin{aligned} \alpha & =\frac{\pi}{4} \text { in }\left(0, \frac{\pi}{2}\right) \\ \therefore \quad \Delta_1 & =2(\sin \alpha) \times 0=0, \end{aligned}$$α∴Δ1​​=4π​ in (0,2π​)=2(sinα)×0=0,​ since value of $\sin \alpha$sinα is finite for $\alpha \in\left(0, \frac{\pi}{2}\right)$α∈(0,2π​) Hence non-trivivial solution for only one value of $\alpha$α in $\left(0, \frac{\pi}{2}\right)$(0,2π​) $$\begin{aligned} & \left|\begin{array}{ccc} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{ccc} 0 & \sin \alpha & \cos \alpha \\ 0 & \cos \alpha & \sin \alpha \\ 2 \cos \alpha & -\sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & \Rightarrow 2 \cos \alpha\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \therefore \cos \alpha=0 \text { or } \sin ^2 \alpha-\cos ^2 \alpha=0 \end{aligned}$$​​cosαsinαcosα​sinαcosα−sinα​cosαsinα−cosα​​=0⇒​002cosα​sinαcosα−sinα​cosαsinα−cosα​​=0⇒2cosα(sin2α−cos2α)=0∴cosα=0 or sin2α−cos2α=0​ But $\cos \alpha=0$cosα=0 not possible for any value of $$\begin{aligned} & \alpha \in\left(0, \frac{\pi}{2}\right) \\ & \therefore \sin ^2 \alpha-\cos ^2 \alpha=0 \Rightarrow \sin \alpha=-\cos \alpha, \end{aligned}$$​α∈(0,2π​)∴sin2α−cos2α=0⇒sinα=−cosα,​ which is also not possible for any value of $$\alpha \in\left(0, \frac{\pi}{2}\right)$$α∈(0,2π​) Hence, there is no solution.