Question

If $y=\cos ax,\text{then}\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|=$ (where $y_n=\frac{d^{n}y}{d{x}^n}$)

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

Given, $y=\cos \left(ax\right)$
Then, $y_1=-a\sin \left(ax\right)=a\cos (\frac{\pi }{2}+ax)$
$y_2=-a^2\cos \left(ax\right)=a^2\cos (\frac{2\pi }{2}+ax)$
$y_3=+a^3\sin \left(ax\right)=a^3\cos (\frac{3\pi }{2}+ax)$
.
.
$y_n=a^n\cos (\frac{n\pi }{2}+ax)$

$\therefore$ The determinant $\Delta \left(x\right)=\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$


$\Delta \left(x\right)=\left|\begin{array}{ccc}\cos ax& a\cos (\frac{\pi }{2}+ax)& a^2\cos (\frac{2\pi }{2}+ax)\\ a^3\cos (\frac{3\pi }{2}+ax)& a^4\cos (\frac{4\pi }{2}+ax)& a^5\cos (\frac{5\pi }{2}+ax)\\ a^6\cos (\frac{6\pi }{2}+ax)& a^7\cos (\frac{7\pi }{2}+ax)& a^8\cos (\frac{8\pi }{2}+ax)\end{array}\right|$

$\Delta \left(x\right)=a^3\times a^6\left|\begin{array}{ccc}\cos \left(ax\right)& -a\sin \left(ax\right)& -a^2\cos \left(ax\right)\\ \sin \left(ax\right)& a\cos \left(ax\right)& -a^2\sin \left(ax\right)\\ -\cos \left(ax\right)& a\sin \left(ax\right)& a^2\cos \left(ax\right)\end{array}\right|$

$R_1\to R_1+R_3$
This gives $\Delta \left(x\right)=a^9\times \left(0\right)$

Hence, $\Delta \left(x\right)=0$