Question

The arbitrary constant on which the value of the determinant $\left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \cos (p-d)a& \cos pa& \cos (p+d)a\\ \sin (p-d)a& \sin pa& \sin (p+d)a\end{array}\right|$ does not depend is

• A. $\alpha$
• B. $p$
• C. $d$
• D. $a$

Answer 1

Answer: (B)

$p$

$\Delta =\left|\begin{array}{ccc}1& \alpha & {\alpha }^2\\ \cos (p-d)a& \cos pa& \cos (p+d)a\\ \sin (p-d)a& \sin pa& \sin (p+d)a\end{array}\right|$

Applying $C_1\to C_1+C_3$

$\Delta =\left|\begin{array}{ccc}1+{\alpha }^2& \alpha & {\alpha }^2\\ \cos (p-d)a+\cos (p+d)a& \cos pa& \cos (p+d)a\\ \sin (p-d)a+\sin (p+d)a& \sin pa& \sin (p+d)a\end{array}\right|$

$=\left|\begin{array}{ccc}1+{\alpha }^2& \alpha & {\alpha }^2\\ 2\cos pa\cos da& \cos pa& \cos (p+d)a\\ 2\sin pa\cos da& \sin pa& \sin (p+d)a\end{array}\right|$

Applying $C_1\to C_1-2\cos da{C}_2$

$\Delta =\left|\begin{array}{ccc}1+{\alpha }^2-2\alpha \cos da& \alpha & {\alpha }^2\\ 0& \cos pa& \cos (p+d)a\\ 0& \sin pa& \sin (p+d)a\end{array}\right|$

$=(1+{\alpha }^2-2\alpha \cos da)\left(\sin \right(p+d)a\cos pa-\cos (p+d\left)a\sin pa\right)$

$=(1+{\alpha }^2-2\alpha \cos da)\sin da$

It does not depend upon the $p$.

Hence, option 'B' is correct.