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

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

Question

(B) Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ \omega & \omega^2 & 1 \ \omega^2 & 1 & \omega \end{array} ight|$.Applying the

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ \omega & \omega^2 & 1 \ \omega^2 & 1 & \omega \end{array} ight|$.Applying the column operation $C_1 	o C_1 + C_2 + C_3$:$\Delta = \left| \begin{array}{ccc} 1 + \omega + \omega^2 & \omega & \omega^2 \ \omega + \omega^2 + 1 & \omega^2 & 1 \ \omega^2 + 1 + \omega & 1 & \omega \end{array} ight|$.Since $1 + \omega + \omega^2 = 0$ for the cube roots of unity,we have:$\Delta = \left| \begin{array}{ccc} 0 & \omega & \omega^2 \ 0 & \omega^2 & 1 \ 0 & 1 & \omega \end{array} ight|$.Since all elements in the first column are $0$,the value of the determinant is $0$.

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

Similar Questions

$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{array} \right|$ where $a = i, b = \omega, c = \omega^2$,then $\Delta$ is equal to

Solve the equation $\left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0$,where $a \neq 0$.

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

If $a, b, c$ are sides of $\triangle ABC$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$,then $\sin^2 A + \sin^2 B + \sin^2 C = $ . . . . . . .

The solutions of the equation $\left| \begin{array}{ccc} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{array} \right| = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module