If the system of linear equations (sin theta) | vedclass

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

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix $A$ must be zero,i.e.,$|A| = 0$.Given the system:$(\sin 	heta) x - y + z = 0$$x - (\cos 	heta) y + z = 0$$x + y + (\sin 	heta) z = 0$The determinant is:$|A| = \begin{vmatrix} \sin 	heta & -1 & 1 \ 1 & -\cos 	heta & 1 \ 1 & 1 & \sin 	heta \end{vmatrix} = 0$Expanding along the first row:$\sin 	heta (-\cos 	heta \cdot \sin 	heta - 1) - (-1) (\sin 	heta - 1) + 1 (1 - (-\cos 	heta)) = 0$$-\sin^2 	heta \cos 	heta - \sin 	heta + \sin 	heta - 1 + 1 + \cos 	heta = 0$$-\sin^2 	heta \cos 	heta + \cos 	heta = 0$$\cos 	heta (1 - \sin^2 	heta) = 0$$\cos 	heta (\cos^2 	heta) = 0$$\cos^3 	heta = 0$$\cos 	heta = 0$For the least positive value of $	heta$,$	heta = \frac{\pi}{2}$.

If the system of linear equations $(\sin \theta) x - y + z = 0$,$x - (\cos \theta) y + z = 0$,and $x + y + (\sin \theta) z = 0$ has a non-trivial solution,then the least positive value of $\theta$ is

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