Question

The value of the determinant
$\begin{vmatrix}1&a&a^{2}\\ cos\left(n-1\right)x&cosnx&cos\left(n+1\right)x\\ sin\left(n+1\right)x&sinnx&sin\left(n+1\right)x\end{vmatrix}$ is zero, if

• A. sin x = 0
• B. cos x = 0
• C. a = 0
• D. $cos\,x = \frac{1+a^{2}}{2a}$

Answer 1

Answer: (A)

Given determinant
$\begin{vmatrix}1&a&a^{2}\\ cos\left(n-1\right)x&cosnx&cos\left(n+1\right)x\\ sin\left(n+1\right)x&sinnx&sin\left(n+1\right)x\end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix}1+a^{2}-2a\,cos\,x&a&a^{2}\\ 0&cosnx&cos\left(n+1\right)x\\ 0&sinnx&sin\left(n+1\right)x\end{vmatrix} = 0$
By applying $C_{1} \to C_{1} + C_{3} - 2\, cos\, x C_{2}$
By expanding
$\left(1 + a^{2} - 2a\, cos\, x\right) \left[cos \,nx \,sin \left(n + 1\right) x- sin \,nx \,\,cos \left(n + 1\right) x\right]= 0$
Now, $\left(1 + a^{2} - 2a\, cos\, x\right)\, sin\, \left(n + 1 - n\right) x = 0$
$\Rightarrow \left(1+ a^{2} -2a\,cos\, x\right)sin \,x = 0$
$sin \,x = 0$ or $cos\, x = \frac{1+a^{2}}{2a}$
As a $\ne 1\quad\therefore \left(\frac{1+a^{2}}{2a}\right) > 1$
$\Rightarrow cos\, x > 1$ It is not possible.
$\therefore sin \,x = 0$