The values of theta, lambda for which the fol | vedclass

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(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The determinant

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$	heta = (2n + 1)\frac{\pi}{2}, \lambda \in \mathbb{R}, n \in \mathbb{I}$

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(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.The determinant is given by:$\Delta = \begin{vmatrix} \sin 	heta & -\cos 	heta & \lambda + 1 \ \cos 	heta & \sin 	heta & -\lambda \ \lambda & \lambda + 1 & \cos 	heta \end{vmatrix} = 0$Expanding the determinant along the first row:$\Delta = \sin 	heta (\sin 	heta \cos 	heta - (-\lambda)(\lambda + 1)) - (-\cos 	heta)(\cos^2 	heta - (-\lambda)(\lambda)) + (\lambda + 1)(\cos 	heta (\lambda + 1) - \lambda \sin 	heta) = 0$Simplifying the expression:$\Delta = \sin 	heta (\sin 	heta \cos 	heta + \lambda^2 + \lambda) + \cos 	heta (\cos^2 	heta + \lambda^2) + (\lambda + 1)(\lambda \cos 	heta + \cos 	heta - \lambda \sin 	heta) = 0$After algebraic simplification,the equation reduces to:$2 \cos 	heta (\lambda^2 + \lambda + 1) = 0$Since $\lambda^2 + \lambda + 1 = (\lambda + \frac{1}{2})^2 + \frac{3}{4} > 0$ for all $\lambda \in \mathbb{R}$,we must have:$\cos 	heta = 0$Therefore,$	heta = (2n + 1)\frac{\pi}{2}$ for $n \in \mathbb{I}$.

The values of $\theta, \lambda$ for which the following equations $\sin \theta x - \cos \theta y + (\lambda + 1)z = 0$; $\cos \theta x + \sin \theta y - \lambda z = 0$; $\lambda x + (\lambda + 1)y + \cos \theta z = 0$ have a non-trivial solution are:

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$\left| {\begin{array}{ccc} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{array}} \right| = $

The number of real values of $t$ such that the system of homogeneous equations
$\begin{aligned}
t x+(t+1) y+(t-1) z &=0 \\
(t+1) x+t y+(t+2) z &=0 \\
(t-1) x+(t+2) y+t z &=0
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has non-trivial solutions is

The set of all values of $\lambda$ for which the system of linear equations $2x_1 - 2x_2 + x_3 = \lambda x_1$,$2x_1 - 3x_2 + 2x_3 = \lambda x_2$,and $-x_1 + 2x_2 = \lambda x_3$ has a non-trivial solution:

If $D = \left|\begin{array}{ccc}1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1\end{array}\right|$,and $p$ and $q$ are the maximum and minimum values of $D$ respectively,then the value of $2p + 3q$ is . . . . . . .

$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$ (in $!$)

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