Question

If $\omega \ne 1$ is cube root of unity, and $A=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right]$ is

• A. symmetric
• B. skew symmetric
• C. singular
• D. orthogonal

Answer 1

Answer: (A), (C)

The correct options are
A symmetric
C singular

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

$A^T=\left[\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right]$

$\Rightarrow A^T=A$

i.e $A$ is a symmetric matrix.

$|A|=\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|=0$

i.e $A$ is singular.

Hence, options A and C.