Question

The determinant \(\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\) is :

• A. Independent of θ only
• B. Independent of x only
• C. Independent of both θ and x
• D. Independent of x but not of θ

Answer 1

The Correct Option is A

To find the determinant of the matrix \(A = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}\), we will use the concept of cofactor expansion along the first row:

\[ \text{det}(A) = x\begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} - \sin\theta\begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} + \cos\theta\begin{vmatrix} -\sin\theta & -x \\ \cos\theta & 1 \end{vmatrix} \]

Calculate each of these 2x2 determinants:

\[ \begin{vmatrix} -x & 1 \\ 1 & x \end{vmatrix} = (-x)(x) - (1)(1) = -x^2 - 1 \]

\[ \begin{vmatrix} -\sin\theta & 1 \\ \cos\theta & x \end{vmatrix} = (-\sin\theta)(x) - (1)(\cos\theta) = -x\sin\theta - \cos\theta \]

\[ \begin{vmatrix} -\sin\theta & -x \\ \cos\theta & 1 \end{vmatrix} = (-\sin\theta)(1) - (-x)(\cos\theta) = -\sin\theta + x\cos\theta \]

Substitute these results back into the original expression:

\[ \text{det}(A) = x(-x^2-1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta) \]

Expanding and simplifying each part yields:

\[ \text{det}(A) = -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \cos\theta\sin\theta + x\cos^2\theta \]

Notice the trigonometric identities \(\sin^2\theta + \cos^2\theta = 1\). Substituting these into the expression gives:

\[ \text{det}(A) = -x^3 - x + x(\sin^2\theta + \cos^2\theta) = -x^3 - x + x(1) = -x^3 \]

The determinant simplifies to \(-x^3\), which is independent of \(\theta\) and depends solely on \(x\). Hence, the determinant is independent of \(\theta\) only.