Matrices & Determinants
Special Determinants
Grade 12
Question:
<p>If \(f(x) = \begin{vmatrix} \cos(x-\phi) & \cos(x-\psi) & \cos(x-\omega) \\ \sin(x-\phi) & \sin(x-\psi) & \sin(x-\omega) \\ \sin(\psi - \omega) & \sin(\omega - \phi) & \sin(\phi - \psi) \end{vmatrix}\), then \(f(9) - 2f(6) + f(3)\) is equal to</p>
<p>(a) 0</p>
<p>(b) \(\phi - \psi\)</p>
<p>(c) \(\phi + \psi + \omega\)</p>
<p>(d) \(\phi + \psi - \omega\)</p>
Step-by-Step Solution
Key Concept: Recognize that the trigonometric structure makes $f(x)$ either constant or linear, so finite differences vanish.
<p>The determinant $f(x)$ represents a trigonometric identity that is constant or a quadratic function in $x$. The second difference $f(9) - 2f(6) + f(3)$ equals twice the second derivative of $f$, which evaluates to 0 for constant or linear functions. By the structure of the determinant (rows 1-2 encode angle and rows become linearly dependent after expansion), $f(x)$ reduces to a constant, giving 0.</p>
Correct Answer: A