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प्रश्न
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
विकल्प
\[Det \left( A \right) = 0\]
\[Det \left( A \right) \in \left( 2, \infty \right)\]
\[Det \left( A \right) \in \left( 2, 4 \right)\]
\[Det \left( A \right) \in \left[ 2, 4 \right]\]
उत्तर
\[\text{ Det }\left( A \right) \in \left[ 2, 4 \right]\]
\[\begin{vmatrix} 1 &\sin \theta & 1\\ - \sin \theta & 1 &\sin \theta\\ - 1 & - \sin \theta & 1 \end{vmatrix}\]
\[ = \begin{vmatrix} 1 & \sin \theta & 2\\ - \sin \theta & 1 & 0 \\- 1 & - \sin \theta & 0 \end{vmatrix} \left[\text{ Applying }C_3 \to C_3 + C_1 \right]\]
\[ = 2 \times \begin{vmatrix} - \sin \theta & 1 \\ - 1 & - \sin \theta \end{vmatrix} \left[\text{ Expanding along }C_3 \right]\]
\[ = 2 \left( \sin^2 \theta + 1 \right)\]
\[\text{ Given: }0 \leq \theta \leq 2\pi \]
\[ \Rightarrow - 1 \leq \sin \theta \leq 1\]
\[ \Rightarrow 0 \leq \sin^2 \theta \leq 1\]
\[\left| A \right| = 2\left( \sin^2 \theta + 1 \right)\]
\[\left| A \right| = 2 \times 1 = 2 \left[ \theta = 0 \right]\]
\[ = 2 \times 2 = 4 \left[ \theta = 2\pi \right]\]
\[ \Rightarrow Det \left( A \right) \in \left[ 2, 4 \right] \]
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