Determinants Matrices Question 44
Question: If $ \omega $ is the cube root of unity, then what is one root of the equation $ \begin{vmatrix} x^{2} & -2x & -2{{\omega }^{2}} \\ 2 & \omega & -\omega \\ 0 & \omega & 1 \\ \end{vmatrix}=0- $
Options:
A) $ 1 $
B) $ -2 $
C) $ 2 $
D) $ \omega $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given matrix is: $ \begin{vmatrix} x^{2} & -2x & -2{{\omega }^{2}} \\ 2 & \omega & -\omega \\ 0 & \omega & 1 \\ \end{vmatrix}=0 $ By $ C_2\to C_2+C_3, $ we get
$ \Rightarrow \begin{vmatrix} x^{2} & -2x-2{{\omega }^{2}} & -2{{\omega }^{2}} \\ 2 & 0 & -\omega \\ 0 & 1+\omega & 1 \\ \end{vmatrix}=0 $
$ \Rightarrow \begin{vmatrix} x^{2} & -2x-2{{\omega }^{2}} & -2{{\omega }^{2}} \\ 2 & 0 & -\omega \\ 0 & -{{\omega }^{2}} & 1 \\ \end{vmatrix}=0 $
$ [\because 1+\omega =-{{\omega }^{2}}] $
$ \Rightarrow {{\omega }^{2}} \begin{vmatrix} x^{2} & -2{{\omega }^{2}} \\ 2 & -\omega \\ \end{vmatrix}+1 \begin{vmatrix} x^{2} & -2x-2{{\omega }^{2}} \\ 2 & -0 \\ \end{vmatrix}=0 $
$ \Rightarrow {{\omega }^{2}}(-\omega x^{2}+4{{\omega }^{2}})-(-4x-4{{\omega }^{2}})=0 $
$ \Rightarrow -x^{2}+4\omega +4x+4{{\omega }^{2}}=0 $
$ \Rightarrow -x^{2}+4\omega -4x-4-4\omega =0 $
$ \Rightarrow -x^{2}-4x-4=0 $
$ \Rightarrow {{(x+2)}^{2}}=0\Rightarrow x=-2 $
BETA
