Let alpha and beta be the roots of the equati | vedclass

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

Question

(C) The roots of the equation $x^2 + x + 1 = 0$ are $\alpha = \omega$ and $\beta = \omega^2,$ where $\omega$ and $\omega^2$ are the complex cube roots

Vedclass · Accepted Answer

$y^3$

Vedclass · Answer

(C) The roots of the equation $x^2 + x + 1 = 0$ are $\alpha = \omega$ and $\beta = \omega^2,$ where $\omega$ and $\omega^2$ are the complex cube roots of unity. Note that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1.$Let the determinant be $\Delta = \left| \begin{array}{ccc} y + 1 & \omega & \omega^2 \ \omega & y + \omega^2 & 1 \ \omega^2 & 1 & y + \omega \end{array} ight|.$Applying the row operation $R_1 	o R_1 + R_2 + R_3$:$\Delta = \left| \begin{array}{ccc} y + 1 + \omega + \omega^2 & y + 1 + \omega + \omega^2 & y + 1 + \omega + \omega^2 \ \omega & y + \omega^2 & 1 \ \omega^2 & 1 & y + \omega \end{array} ight|$Since $1 + \omega + \omega^2 = 0,$ the first row becomes $y, y, y.$$\Delta = y \left| \begin{array}{ccc} 1 & 1 & 1 \ \omega & y + \omega^2 & 1 \ \omega^2 & 1 & y + \omega \end{array} ight|$Applying column operations $C_2 	o C_2 - C_1$ and $C_3 	o C_3 - C_1$:$\Delta = y \left| \begin{array}{ccc} 1 & 0 & 0 \ \omega & y + \omega^2 - \omega & 1 - \omega \ \omega^2 & 1 - \omega^2 & y + \omega - \omega^2 \end{array} ight|$Expanding along the first row:$\Delta = y \left[ (y + \omega^2 - \omega)(y + \omega - \omega^2) - (1 - \omega)(1 - \omega^2) ight]$$\Delta = y \left[ (y + (\omega^2 - \omega))(y - (\omega^2 - \omega)) - (1 - \omega^2 - \omega + \omega^3) ight]$$\Delta = y \left[ y^2 - (\omega^2 - \omega)^2 - (1 - (\omega^2 + \omega) + 1) ight]$Since $\omega^2 + \omega = -1,$ we have $\Delta = y \left[ y^2 - (\omega^4 - 2\omega^3 + \omega^2) - (1 - (-1) + 1) ight] = y \left[ y^2 - (\omega - 2 + \omega^2) - 3 ight]$$\Delta = y \left[ y^2 - (-1 - 2) - 3 ight] = y \left[ y^2 + 3 - 3 ight] = y^3.$Thus,the value is $y^3.$

Let $\alpha$ and $\beta$ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $\mathbb{R},$ the determinant $\left| \begin{array}{ccc} y + 1 & \alpha & \beta \\ \alpha & y + \beta & 1 \\ \beta & 1 & y + \alpha \end{array} \right|$ is equal to

Similar Questions

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 $t$.

Let $a, b, c > 0$ and $\Delta = \begin{vmatrix} a+b & b & c \\ b+c & c & a \\ c+a & a & b \end{vmatrix}$. Then which of the following is not correct?

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

Find the values of $k$ if the area of the triangle is $4$ square units and the vertices are $(-2, 0), (0, 4), (0, k)$.

If $A = \begin{bmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{bmatrix}$ is a singular matrix,then the possible values of $x$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module