If matrix A = begin{bmatrix} sin theta & csc | vedclass

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(D) matrix is non-invertible if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} \sin 	heta & \csc 	heta & 1 \ \csc 	heta & 1 & \

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$n\pi + (-1)^n \frac{\pi}{6}$

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(D) matrix is non-invertible if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} \sin 	heta & \csc 	heta & 1 \ \csc 	heta & 1 & \sin 	heta \ 1 & \sin 	heta & \csc 	heta \end{bmatrix}$.Calculating the determinant $|A| = \sin 	heta (\csc 	heta \cdot \csc 	heta - \sin^2 	heta) - \csc 	heta (\csc^2 	heta - \sin 	heta) + 1 (\sin 	heta \csc 	heta - 1) = 0$.$|A| = \sin 	heta \csc^2 	heta - \sin^3 	heta - \csc^3 	heta + \sin 	heta \csc 	heta + 1 - 1 = 0$.$|A| = \csc 	heta - \sin^3 	heta - \csc^3 	heta + 1 = 0$.Let $x = \sin 	heta$. Then $\csc 	heta = \frac{1}{x}$.$\frac{1}{x} - x^3 - \frac{1}{x^3} + 1 = 0 \Rightarrow \frac{x^2 - x^6 - 1 + x^3}{x^3} = 0$.$x^6 - x^3 - x^2 + 1 = 0 \Rightarrow (x^3 - 1)(x^3 - x^2 + 1) = 0$.Since $x^3 - x^2 + 1 = 0$ has no real roots for $\sin 	heta$,we have $x^3 = 1 \Rightarrow \sin 	heta = 1$.Thus,$	heta = 2n\pi + \frac{\pi}{2}$.

If matrix $A = \begin{bmatrix} \sin \theta & \csc \theta & 1 \\ \csc \theta & 1 & \sin \theta \\ 1 & \sin \theta & \csc \theta \end{bmatrix}$ is a non-invertible matrix,then the possible value of $\theta$ is $(n \in \mathbb{Z})$

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