The system of linear equations (sin theta) x | vedclass

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(A) For a system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.Let $D = \begi

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no value of $	heta$

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(A) For a system of homogeneous linear equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero.Let $D = \begin{vmatrix} \sin 	heta & 1 & -2 \ 2 & -1 & \cos 	heta \ -3 & \sec 	heta & 3 \end{vmatrix} = 0$.Expanding along the first row:$D = \sin 	heta (-3 - \cos 	heta \sec 	heta) - 1(6 + 3 \cos 	heta) - 2(2 \sec 	heta - 3) = 0$.Since $\cos 	heta \sec 	heta = 1$, we have:$D = \sin 	heta (-3 - 1) - 6 - 3 \cos 	heta - 4 \sec 	heta + 6 = 0$.$-4 \sin 	heta - 3 \cos 	heta - 4 \sec 	heta = 0$.Multiply by $\cos 	heta$ (given $	heta 
eq (2n+1)\frac{\pi}{2}$, so $\cos 	heta 
eq 0$):$-4 \sin 	heta \cos 	heta - 3 \cos^2 	heta - 4 = 0$.$-2 \sin(2	heta) - 3 \cos^2 	heta - 4 = 0$.Using $\cos^2 	heta = \frac{1 + \cos(2	heta)}{2}$:$-2 \sin(2	heta) - \frac{3}{2} - \frac{3}{2} \cos(2	heta) - 4 = 0$.$-4 \sin(2	heta) - 3 \cos(2	heta) = 11$.Since the maximum value of $a \sin x + b \cos x$ is $\sqrt{a^2 + b^2} = \sqrt{(-4)^2 + (-3)^2} = 5$, and $5 Thus, the system has no non-trivial solution for any $	heta$.

The system of linear equations $(\sin \theta) x + y - 2z = 0$, $2x - y + (\cos \theta) z = 0$, and $-3x + (\sec \theta) y + 3z = 0$, where $\theta \neq (2n + 1) \frac{\pi}{2}$, has a non-trivial solution for:

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