Statement -1: The system of linear equationsx | vedclass

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(B) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.For

Vedclass · Accepted Answer

Statement $-1$ is true,Statement $-2$ is true,Statement $-2$ is a correct explanation for Statement $-1$.

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(B) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.For Statement $-1$,the determinant is:$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ 1 & -\sin \alpha & -\cos \alpha \end{matrix} ight|$Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \ 0 & \cos \alpha - \sin \alpha & \sin \alpha - \cos \alpha \ 0 & -2\sin \alpha & -2\cos \alpha \end{matrix} ight|$$= 1 \cdot [(\cos \alpha - \sin \alpha)(-2\cos \alpha) - (\sin \alpha - \cos \alpha)(-2\sin \alpha)]$$= -2\cos^2 \alpha + 2\sin \alpha \cos \alpha + 2\sin^2 \alpha - 2\sin \alpha \cos \alpha = 2(\sin^2 \alpha - \cos^2 \alpha) = -2\cos(2\alpha)$.Setting $\Delta_1 = 0$,we get $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$2\alpha \in (0, \pi)$,so $2\alpha = \frac{\pi}{2} \Rightarrow \alpha = \frac{\pi}{4}$. Thus,Statement $-1$ is true.For Statement $-2$,the determinant is:$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \ \sin \alpha & \cos \alpha & \sin \alpha \ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} ight|$Applying $C_3 	o C_3 - C_1$:$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & 0 \ \sin \alpha & \cos \alpha & 0 \ \cos \alpha & -\sin \alpha & -2\cos \alpha \end{matrix} ight|$$= -2\cos \alpha (\cos^2 \alpha - \sin^2 \alpha) = -2\cos \alpha \cos(2\alpha) = 0$.This implies $\cos \alpha = 0$ or $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$\cos \alpha 
eq 0$ and $\cos(2\alpha) = 0$ only at $\alpha = \frac{\pi}{4}$. Thus,Statement $-2$ is true and explains Statement $-1$.

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

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