The system of equations lambda x - y + (costh | vedclass

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

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for no value of $\lambda$ and $	heta$

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(A) For a system of homogeneous linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.The coefficient matrix is:$A = \begin{bmatrix} \lambda & -1 & \cos	heta \ 3 & 1 & 2 \ \cos	heta & 1 & 2 \end{bmatrix}$Calculating the determinant $|A|$:$|A| = \lambda(1 	imes 2 - 1 	imes 2) - (-1)(3 	imes 2 - \cos	heta 	imes 2) + \cos	heta(3 	imes 1 - \cos	heta 	imes 1)$$|A| = \lambda(0) + (6 - 2\cos	heta) + \cos	heta(3 - \cos	heta)$$|A| = 6 - 2\cos	heta + 3\cos	heta - \cos^2	heta$$|A| = 6 + \cos	heta - \cos^2	heta$For a non-trivial solution,$|A| = 0$:$-\cos^2	heta + \cos	heta + 6 = 0$$\cos^2	heta - \cos	heta - 6 = 0$Let $t = \cos	heta$. Then $t^2 - t - 6 = 0$.$(t - 3)(t + 2) = 0$So,$\cos	heta = 3$ or $\cos	heta = -2$.Since the range of $\cos	heta$ is $[-1, 1]$,neither $\cos	heta = 3$ nor $\cos	heta = -2$ is possible for any real $	heta$.Therefore,$|A|$ is never zero for any $	heta \in (0, 2\pi)$.Thus,the system has only the trivial solution $(x=0, y=0, z=0)$ for all values of $\lambda$ and $	heta$.

The system of equations $\lambda x - y + (\cos\theta) z = 0$,$3x + y + 2z = 0$,and $(\cos\theta) x + y + 2z = 0$ for $0 < \theta < 2\pi$ has non-trivial solution$(s)$:

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