If n ne 3k and 1, omega, omega^2 are the cube | vedclass

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

Question

(A) Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \begin{vmatrix} 1 + \omega^n + \omega^{2n} & \omega^n & \omega^{2n} \ 1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \begin{vmatrix} 1 + \omega^n + \omega^{2n} & \omega^n & \omega^{2n} \ 1 + \omega^n + \omega^{2n} & 1 & \omega^n \ 1 + \omega^n + \omega^{2n} & \omega^{2n} & 1 \end{vmatrix}$Since $n 
e 3k$,$n$ is not a multiple of $3$. Therefore,$1 + \omega^n + \omega^{2n} = 0$.Substituting this into the determinant:$\Delta = \begin{vmatrix} 0 & \omega^n & \omega^{2n} \ 0 & 1 & \omega^n \ 0 & \omega^{2n} & 1 \end{vmatrix}$Since all elements in the first column are $0$,the value of the determinant is $0$.

If $n \ne 3k$ and $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1 \end{vmatrix}$ has the value

Similar Questions

Which of the following is correct?

The matrix $\begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$ is not invertible if $a$ has the value:

The value of the determinant $\left| \begin{array}{ccc} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{array} \right|$ is

Find the area of the triangle with vertices at the points $(-2, -3), (3, 2), (-1, -8)$.

The area of a triangle is $13$ sq. units whose vertices are $A(8, 2)$,$B(k, 4)$,and $C(6, 7)$. Then,the integer value of $k$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module