Algebraic Twists!

Algebra Level 5

$f(x) = \begin{bmatrix} \sin(x+\alpha) & \sin(x+\beta) & \sin(x+\gamma) \\ \cos(x+\alpha) & \cos(x+\beta) & \cos(x+\gamma) \\ \cos(\beta-\gamma) & \cos(\gamma-\alpha) & \cos(\alpha-\beta) \end{bmatrix}$

If $f(x)$ is as defined above and $f(9)=\lambda \ne 0$ then let $\displaystyle m=\frac{\sum_{k=1}^{9}f(k)}{f(9)}$ ,

$P= \begin{bmatrix}\cos\frac{\pi}{9} & \sin\frac{\pi}{9} \\ -\sin\frac{\pi}{9} & \cos\frac{\pi}{9} \end{bmatrix}$

and $\alpha$ , $\beta$ and $\gamma$ be non-real numbers such that $(\alpha P^6+\beta P^3+\gamma I)$ is the zero matrix. Then let the value of $(\alpha^2+\beta^2+\gamma^2)^{(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)}$ be $n$ .

Evaluate $m-n$

The answer is 8.00.

1 solution

Rajdeep Brahma
Jul 5, 2018

$A:$ It is easy to see that $f'(x)=0$ .So $f(x)=\lambda$ and the answer is $9$ .

$B:$ Note that $\begin{bmatrix}{cos\frac{\pi}{9}} && {sin\frac{\pi}{9}} \\ {-sin\frac{\pi}{9}} && {cos\frac{\pi}{9}}\\ \end{bmatrix}$ is a rotation unit vector making an angle of $\frac{\pi}{9}$ with the $x-axis$ .So $P^n$ makes an angle of $\frac{n\pi}{9}$ with the $x-axis$ .Now the sum of all the unit vectors $\alphap^6,\betap^3,\gamma*I$ is $0$ ....(very similar to addition of $1,\omega,\omega^2)$ .Now use the triangle law of addition to deduce that $\alpha=-\beta=\gamma$ and so the answer will be $1$ .

Ki level. Haha. Btw. Bhalo qs. Moja laglo. I used Characteristic equation in 2nd one. And in 1st one I found out it as a product of two matrix and one of them was const and other ones Det was 1 so. 9/1.