Mathematics - Determinant Question with Solution | TestHub

Applying ${\mathrm{C}}_1\to {\mathrm{C}}_1-\sin \theta {\mathrm{C}}_3$ and ${\mathrm{C}}_2\to {\mathrm{C}}_2+\cos \theta {\mathrm{C}}_3$, we get:

$$f(\theta )=\left|\begin{array}{ccc}1& 0& -\sin \theta \\ 0& 1& \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right|$$

Applying ${\mathrm{R}}_3\to {\mathrm{R}}_3-\sin \theta {\mathrm{R}}_1+\cos \theta {\mathrm{R}}_2$, we get:

$$f(\theta )=\left|\begin{array}{ccc}1& 0& -\sin \theta \\ 0& 1& \cos \theta \\ 0& 0& {\sin }^2\theta +{\cos }^2\theta \end{array}\right|=1$$

Thus, $\mathrm{f}\left(\frac{\pi }{6}\right)=1$.