Question

If $y=\sin mx$ then the value of the determinant $\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|,$ where $y_n=\frac{d^{n}y}{d{x}^n},$ is

• A. $m^9$
• B. $m^2$
• C. $m^3$
• D. None of the above

Answer 1

The correct option is D None of the above

We have $y=\sin mx$,
$\therefore y_1=m\cos mx,$
$y_2=-m^2\sin mx,$
$...$
$y_n=m^n\sin (\frac{n\pi }{2}+mx)$

$\therefore △=\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$

$=\left|\begin{array}{ccc}\sin mx& m\cos mx& -m^2\sin mx\\ -m^3\cos mx& m^4\sin mx& m^5\cos mx\\ -m^6\sin mx& -m^7\cos mx& m^8\sin mx\end{array}\right|$

$=m^{12}\left|\begin{array}{ccc}\sin mx& \cos mx& -\sin mx\\ -\cos mx& \sin mx& \cos mx\\ -\sin mx& -\cos mx& \sin mx\end{array}\right|$

$R_1\to R_1+R_3$

$=m^{12}\left|\begin{array}{ccc}0& 0& 0\\ -\cos mx& \sin mx& \cos mx\\ -\sin mx& -\cos mx& \sin mx\end{array}\right|$

$=0$