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प्रश्न
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
उत्तर
\[\begin{vmatrix}\sin^2 23^\circ & \sin^2 67^\circ & \cos180^\circ \\ - \sin^2 67^\circ & - \sin^2 23^\circ & \cos^2 180^\circ \\ \cos180^\circ & \sin^2 23^\circ & \sin^2 67^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ & \sin^2 \left( 90 - 23 \right)^\circ & - 1 \\ - \sin^2 \left( 90 - 23 \right)^\circ & - \sin^2 23^\circ & 1 \\ - 1 & \sin^2 23^\circ & \sin^2 \left( 90 - 23 \right)^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ & \cos^2 23^\circ & - 1 \\ - \cos^2 23^\circ & - \sin^2 23^\circ & 1 \\ - 1 & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = \begin{vmatrix}\sin^2 23^\circ + \cos^2 23^\circ & \cos^2 23^\circ & - 1 \\ - \cos^2 23^\circ - \sin^2 23^\circ & - \sin^2 23^\circ & 1 \\ - 1 + \sin^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix} \left[ \text{ Applying }C_1 \to C_1 + C_2 \right]\]
\[ = \begin{vmatrix}1 & 1 & - 1 \\ - 1 & - \sin^2 23^\circ & 1 \\ - \cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = \left( - 1 \right)\begin{vmatrix}- 1 & 1 & - 1 \\ 1 & - \sin^2 23^\circ & 1 \\ \cos^2 23^\circ & \sin^2 23^\circ & \cos^2 23^\circ\end{vmatrix}\]
\[ = 0\]
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