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Let a = ⎡ ⎢ ⎣ 1 Sin θ 1 − Sin θ 1 Sin θ − 1 − Sin θ 1 ⎤ ⎥ ⎦ , Where 0 ≤ θ ≤ 2 π . Then, (A) D E T ( a ) = 0 (B) D E T ( a ) ∈ ( 2 , ∞ ) (C) D E T ( a ) ∈ ( 2 , 4 ) (D) - Mathematics

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Question

Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]



Options

  • \[Det \left( A \right) = 0\]

  • \[Det \left( A \right) \in \left( 2, \infty \right)\]

  • \[Det \left( A \right) \in \left( 2, 4 \right)\]

  • \[Det \left( A \right) \in \left[ 2, 4 \right]\]

MCQ
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Solution

 \[\text{ Det }\left( A \right) \in \left[ 2, 4 \right]\]

\[\begin{vmatrix} 1 &\sin \theta & 1\\ - \sin \theta & 1 &\sin \theta\\ - 1 & - \sin \theta & 1 \end{vmatrix}\]
\[ = \begin{vmatrix} 1 & \sin \theta & 2\\ - \sin \theta & 1 & 0 \\- 1 & - \sin \theta & 0 \end{vmatrix} \left[\text{ Applying }C_3 \to C_3 + C_1 \right]\]
\[ = 2 \times \begin{vmatrix} - \sin \theta & 1 \\ - 1 & - \sin \theta \end{vmatrix} \left[\text{ Expanding along }C_3 \right]\]

\[ = 2 \left( \sin^2 \theta + 1 \right)\]

\[\text{ Given: }0 \leq \theta \leq 2\pi \]

\[ \Rightarrow - 1 \leq \sin \theta \leq 1\]

\[ \Rightarrow 0 \leq \sin^2 \theta \leq 1\]

\[\left| A \right| = 2\left( \sin^2 \theta + 1 \right)\]

\[\left| A \right| = 2 \times 1 = 2 \left[ \theta = 0 \right]\]

\[ = 2 \times 2 = 4 \left[ \theta = 2\pi \right]\]

\[ \Rightarrow Det \left( A \right) \in \left[ 2, 4 \right] \]

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Chapter 6: Determinants - Exercise 6.7 [Page 95]

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RD Sharma Mathematics [English] Class 12
Chapter 6 Determinants
Exercise 6.7 | Q 21 | Page 95

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