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

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

Question

(D) Given the determinant equation: $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \o

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) Given the determinant equation: $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \omega \end{array} ight| = 0$. Applying the row operation $R_1 ightarrow R_1 + R_2 + R_3$: The sum of the first row becomes: $(x + 2 + \omega + \omega^2) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1 + 0) = (x + 1)$. Similarly,the other elements in the first row become: $(\omega + x + 1 + \omega^2 + 1) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1)$. $(\omega^2 + 1 + x + 1 + \omega) = (x + 1 + (1 + \omega + \omega^2)) = (x + 1)$. Factoring out $(x + 1)$ from the first row: $(x + 1) \left| \begin{array}{ccc} 1 & 1 & 1 \ \omega & x + 1 + \omega^2 & 1 \ \omega^2 & 1 & x + 1 + \omega \end{array} ight| = 0$. Since the determinant part evaluates to a non-zero value for $x 
eq -1$,the equation simplifies to $(x + 1)^3 = 0$. Therefore,the root is $x = -1$.

If $\omega$ is a cube root of unity,then the root of the equation $\left| \begin{array}{ccc} x + 2 & \omega & \omega^2 \\ \omega & x + 1 + \omega^2 & 1 \\ \omega^2 & 1 & x + 1 + \omega \end{array} \right| = 0$ is:

Similar Questions

Evaluate the determinant: $\left|\begin{array}{cc}x^{2}-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$

Area of the triangle formed by points $(102, -4)$,$(105, -2)$,and $(103, -3)$ is:

If $x, y, z$ are all different and not equal to zero and $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0$,then the value of $x^{-1}+y^{-1}+z^{-1}$ is equal to

The matrix $\begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$ is invertible,if

If $a, b, c$ are real numbers such that $a-b=1$ and $b-c=3$, then the number of matrices of the form $A=\begin{bmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{bmatrix}$ such that $|A|=-12$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module