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

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

Question

(A) Let $\Delta = \left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \ 1 & 1 & 1 \ 1 & -1 & 0 \end{array} ight|$.Expanding along the third row:$\De

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $\Delta = \left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \ 1 & 1 & 1 \ 1 & -1 & 0 \end{array} ight|$.Expanding along the third row:$\Delta = 1 \cdot \left| \begin{array}{cc} 2\omega & -\omega^2 \ 1 & 1 \end{array} ight| - (-1) \cdot \left| \begin{array}{cc} 2 & -\omega^2 \ 1 & 1 \end{array} ight| + 0 \cdot \left| \begin{array}{cc} 2 & 2\omega \ 1 & 1 \end{array} ight|$$\Delta = 1(2\omega - (-\omega^2)) + 1(2 - (-\omega^2))$$\Delta = 2\omega + \omega^2 + 2 + \omega^2$$\Delta = 2\omega + 2\omega^2 + 2$$\Delta = 2(1 + \omega + \omega^2)$Since $\omega$ is a complex cube root of unity,$1 + \omega + \omega^2 = 0$.Therefore,$\Delta = 2(0) = 0$.

If $\omega$ is a complex cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{array} \right|$ is:

Similar Questions

If the matrix $\begin{bmatrix} 1 & 3 & \lambda + 2 \\ 2 & 4 & 8 \\ 3 & 5 & 10 \end{bmatrix}$ is singular,then $\lambda = $

If the system of equations $ax + y + z = 0$,$x + by + z = 0$ and $x + y + cz = 0$,where $a, b, c \neq 1$,has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$, then $\det(A)$ is equal to

If $\left|\begin{array}{cc}x^3+2 x^2+3 x-2 & x^2+2 x+4 \\ x^3-x^2-2 x-1 & 3 x^3-2 x^2+4 x-2\end{array}\right| = a x^6+b x^5+c x^4+d x^3+e x^2+f x+g$,then $a+b+c+d+e+f$ is equal to

If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$,then $x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module