If omega is a cube root of unity,then left| b | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $\Delta = \left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \ \omega & x + \omega^2 & 1 \ \omega^2 & 1 & x + \omega \end{array} ight|$.A

Vedclass · Accepted Answer

$x^3$

Vedclass · Answer

(D) Let $\Delta = \left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \ \omega & x + \omega^2 & 1 \ \omega^2 & 1 & x + \omega \end{array} ight|$.Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \left| \begin{array}{ccc} x + 1 + \omega + \omega^2 & \omega & \omega^2 \ x + 1 + \omega + \omega^2 & x + \omega^2 & 1 \ x + 1 + \omega + \omega^2 & 1 & x + \omega \end{array} ight|$.Since $1 + \omega + \omega^2 = 0$,the first column becomes $x$:$\Delta = x \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ 1 & x + \omega^2 & 1 \ 1 & 1 & x + \omega \end{array} ight|$.Expanding along the first column:$\Delta = x [ 1((x + \omega^2)(x + \omega) - 1) - 1(\omega(x + \omega) - \omega^2) + 1(\omega - \omega^2(x + \omega^2)) ]$.Simplifying the terms inside:$= x [ (x^2 + \omega x + \omega^2 x + \omega^3 - 1) - (\omega x + \omega^2 - \omega^2) + (\omega - \omega^2 x - \omega^4) ]$.Using $\omega^3 = 1$ and $\omega^4 = \omega$:$= x [ x^2 + \omega x + \omega^2 x + 1 - 1 - \omega x + \omega - \omega^2 x - \omega ] = x [ x^2 ] = x^3$.

If $\omega$ is a cube root of unity,then $\left| \begin{array}{ccc} x + 1 & \omega & \omega^2 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{array} \right| = $

Similar Questions

If $\omega$ is an imaginary root of unity,then the value of $\left| \begin{array}{ccc} a & b\omega^2 & a\omega \\ b\omega & c & b\omega^2 \\ c\omega^2 & a\omega & c \end{array} \right|$ is

Let $A = \begin{bmatrix} 3-t & 1 & 0 \\ -1 & 3-t & 1 \\ 0 & -1 & 0 \end{bmatrix}$ and $\det(A) = 5$, then find the value of $t$.

If $D = \left|\begin{array}{ccc}1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1\end{array}\right|$,and $p$ and $q$ are the maximum and minimum values of $D$ respectively,then the value of $2p + 3q$ is . . . . . . .

The cubic equation $\left| \begin{array}{ccc} 0 & a-x & b-x \\ -a-x & 0 & c-x \\ -b-x & -c-x & 0 \end{array} \right| = 0$ has a repeated root in $x$. Then:

$A$ root of the equation $\left| \begin{array}{ccc} 3 - x & -6 & 3 \\ -6 & 3 - x & 3 \\ 3 & 3 & -6 - x \end{array} \right| = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module