Question

If $\omega \ne 1$ is a cube root of unity, then

$A=\left[\begin{array}{ccc}1+2{\omega }^{100}+{\omega }^{200}& {\omega }^2& 1\\ 1& 1+{\omega }^{101}+2{\omega }^{202}& \omega \\ \omega & {\omega }^2& 2+{\omega }^{100}+2{\omega }^{200}\end{array}\right]$

• A. $A$ is singular
• B. $|A|=0$
• C. $A$ is symmetric
• D. none of these

Answer 1

Answer: (A), (B)

The correct options are
A $A$ is singular
B $|A|=0$

$A=\left|\begin{array}{ccc}1+{\omega }^{100}+{\omega }^{200}& {\omega }^2& 1\\ 1& 1+{\omega }^{101}+{\omega }^{202}& \omega \\ \omega & {\omega }^2& 2+{\omega }^{100}+{\omega }^{200}\end{array}\right|$

$1+\omega +{\omega }^2=0$

${\omega }^{100}={\omega }^{99}\times \omega =\omega$

${\omega }^{200}={\omega }^{198}\times {\omega }^2={\omega }^2$

${\omega }^{101}={\omega }^2$${\omega }^{202}={\omega }^4=\omega$

$A=\left|\begin{array}{ccc}1+2\omega +{\omega }^2& {\omega }^2& 1\\ 1& 1+{\omega }^2+2\omega & \omega \\ \omega & {\omega }^2& 2+\omega +2{\omega }^2\end{array}\right|$

$A=\left|\begin{array}{ccc}\omega & {\omega }^2& 1\\ 1& \omega & \omega \\ \omega & {\omega }^2& -\omega \end{array}\right|$

$A=\omega \left|\begin{array}{ccc}\omega & \omega & 1\\ 1& 1& \omega \\ \omega & \omega & -\omega \end{array}\right|$

$C_1\&C_2$are same

$\left|A\right|=0$

$A\to$singular

$△=0$

Option A and B