\( \begin{vmatrix}\lambda & \sin\theta & \cos\theta\\-\sin\theta & -\lambda & 1\\\cos\theta & 1 & \lambda\end{vmatrix} \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( -\lambda^3 \)
Expanding along the first row:
\( \lambda(-\lambda^2-1)-\sin\theta(-\lambda\sin\theta-\cos\theta)+\cos\theta(-\sin\theta+\lambda\cos\theta) \)
\( =-\lambda^3-\lambda+\lambda\sin^2\theta+\sin\theta\cos\theta-\sin\theta\cos\theta+\lambda\cos^2\theta \)
\( =-\lambda^3-\lambda+\lambda(\sin^2\theta+\cos^2\theta) \)
\( =-\lambda^3 \)
Shortcut: The options do not contain \( \theta \), so put \( \theta=0 \):
\( \displaystyle \begin{vmatrix}\lambda & 0 & 1\\0 & -\lambda & 1\\1 & 1 & \lambda\end{vmatrix}=\lambda(-\lambda^2-1)+1(0+\lambda) \)
\( =-\lambda^3 \)