Q. If $y = \sin{px\ and\ y_{n}}$ is the $n$th derivative of $y$, then $\left| \begin{matrix} y & y_{1} & y_{2} \\ y_{3} & y_{4} & y_{5} \\ y_{6} & y_{7} & y_{8} \end{matrix} \right|$ is equal to

A) 1
B) 0
C) $- 1$
D) None of these
Explanation

(b)
Let $D = \left| \begin{matrix}
y & y_{1} & y_{2} \\
y_{3} & y_{4} & y_{5} \\
y_{6} & y_{7} & y_{8}
\end{matrix} \right|$
$\Rightarrow D = \left| \begin{matrix}
\sin{px} & p\cos{px} & - p^{2}\sin{px} \\
- p^{3}\cos{px} & p^{4}\sin{px} & p^{5}\cos{px} \\
- p^{6}\sin{px} & - p^{7}\cos{px} & p^{8}\sin{px}
\end{matrix} \right|$
Taking $p^{3}\ and\ p^{6}$common from $R_{2}\ and\ R_{3}$row
$= p^{9}\left| \begin{matrix}
\sin{px} & p\cos{px} & - p^{2}\sin{px} \\
- \cos{px} & p\sin{px} & p^{2}\cos{px} \\
- \sin{px} & - p\cos{px} & p^{2}\sin{px}
\end{matrix} \right|$
$= - p^{9}\left| \begin{matrix}
\sin{px} & p\cos{px} & - p^{2}\sin{px} \\
- \cos{px} & p\sin{px} & p^{2}\cos{px} \\
\sin{px} & p\cos{px} & - p^{2}\sin{px}
\end{matrix} \right|$
$= 0$ ($R_{1}$and $R_{3}$rows are identical)

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.