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

The values of $x$ for which the given matrix $\left[\begin{array}{ccc}-x & x & 2 \\ 2 & x & -x \\ x & -2 & -2\end{array}\right]$ will be non-singular are

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

If $A_\alpha = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, then the determinant of $A_{\pi / 5} A_{\pi / 4} A_{3 \pi / 10}$ is:

The area of a triangle with vertices $(-3, 0)$,$(3, 0)$,and $(0, k)$ is $9$ sq units. Find the value of $k$.

The area of a triangle with vertices $(K, 0), (4, 0), (0, 2)$ is $4$ square units. Find the value of $K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module