Question

If $y=\cos ax$; 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 $y_{\tau }=-\frac{d^{\tau }}{d{x}^{\tau }}\cdot y.$ Calculate the value of determinant.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

Given, $y=\cos \left(ax\right)$
Then, $y_1=-a\sin \left(ax\right)=a\cos (\frac{\pi }{2}+ax)$
$y_2=-a^2\cos \left(ax\right)=a^2\cos (\frac{2\pi }{2}+ax)$
$y_3=+a^3\sin \left(ax\right)=a^3\cos (\frac{3\pi }{2}+ax)$
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$y_n=a^n\cos (\frac{n\pi }{2}+ax)$
$\therefore$ The determinant $\Delta \left(x\right)=\left|\begin{array}{ccc}y& y_1& y_2\\ y_3& y_4& y_5\\ y_6& y_7& y_8\end{array}\right|$

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

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

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

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