If \(\begin{vmatrix}x & \sin\theta & \cos\theta\\ -\sin\theta & -x & 1\\ \cos\theta & 1 & x\end{vmatrix} = 27\), then the value of x is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-3\)
Expanding along the first row:
\(\Delta = x(-x^{2} - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)\)
\(\Delta = -x^{3} - x + x\sin^{2}\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^{2}\theta\)
Using \(\sin^{2}\theta + \cos^{2}\theta = 1\), the terms in \(\theta\) cancel:
\(\Delta = -x^{3} - x + x\)
\(\Delta = -x^{3}\)
So
\(-x^{3} = 27\)
\(x^{3} = -27\)
\(x = -3\)
Why the other options are incorrect
Option (2) 3 — Incorrect. With \(x = 3\) the determinant is \(-27\), not 27.
Option (3) \(\pm 3\) — Incorrect. \(x^{3} = -27\) has only one real root; \(x = 3\) gives \(-27\).
Option (4) 0 — Incorrect. With \(x = 0\) the determinant is 0.