Question

If $f\left(\theta \right)=\left|\begin{array}{ccc}{\cos }^2\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right|$ then for all $\theta$,

• A. $f\left(\theta \right)=1$
• B. $f\left(\theta \right)=2$
• C. $f\left(\theta \right)=3$
• D. None of the above

Answer 1

Answer: (A)

$f\left(\theta \right)=1$

$f\left(\theta \right)=\left|\begin{array}{ccc}{\cos }^2\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & {\sin }^2\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right|$
$f\left(\theta \right)=\left|\begin{array}{ccc}1& 0& -\sin \theta \\ 0& 1& \cos \theta \\ \sin \theta & -\cos \theta & 0\end{array}\right|$
$\text{From,}{C}_1\to C_1-\sin \theta C_3,C_2\to C_2+\cos \theta C_3$
$=\left|\begin{array}{ccc}1& 0& -\sin \theta \\ 0& 1& \cos \theta \\ 0& -\cos \theta & {\sin }^2\theta \end{array}\right|$
Expand along first column, we get
$={\sin }^2\theta +{\cos }^2\theta$
$\therefore f\left(\theta \right)=1$