If 1, omega, omega^2 are the cube roots of un | vedclass

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

Question

(A) We are given the determinant $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We are given the determinant $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end{vmatrix}$.Expanding the determinant along the first row:$\Delta = 1(\omega^n \cdot \omega^n - 1 \cdot 1) - \omega^n(\omega^n \cdot \omega^n - \omega^{2n} \cdot 1) + \omega^{2n}(\omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n})$$\Delta = 1(\omega^{2n} - 1) - \omega^n(\omega^{2n} - \omega^{2n}) + \omega^{2n}(\omega^n - \omega^{4n})$Since $\omega^3 = 1$,we have $\omega^{3n} = 1$ and $\omega^{4n} = \omega^n$.$\Delta = (\omega^{2n} - 1) - 0 + \omega^{2n}(\omega^n - \omega^n)$$\Delta = \omega^{2n} - 1 + 0 = \omega^{2n} - 1$.However,if we look at the properties of the cube roots of unity,the sum of the columns $C_1 + C_2 + C_3$ gives:$1 + \omega^n + \omega^{2n}$.If $n$ is not a multiple of $3$,$1 + \omega^n + \omega^{2n} = 0$,making the determinant $0$.If $n$ is a multiple of $3$,then $\omega^n = 1$,so the determinant becomes $\begin{vmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{vmatrix} = 0$.Thus,in all cases,$\Delta = 0$.

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$ is equal to

Similar Questions

If $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a + 1}&{a - b}&{b - c} \end{array}} \right| = 0$,where $a, b, c \in R - \{0\}$,then:

$\left| {\begin{array}{ccc} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right| = $

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1\end{array}\right|=1+a^{2}+b^{2}+c^{2}$

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

If $\begin{vmatrix} ^9C_4 & ^9C_5 & ^{10}C_r \\ ^{10}C_6 & ^{10}C_7 & ^{11}C_{r+2} \\ ^{11}C_8 & ^{11}C_9 & ^{12}C_{r+4} \end{vmatrix} = 0$,then $r$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module