Solution:
Given
It is also given that
![\[\omega \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-446c6f033f432dfb8a90a788da7b6ec5_l3.png "Rendered by QuickLaTeX.com")
is a complex cube root of unity,
Consider the LHS,
We know that
![\[1\text{ }+\text{ }\omega \text{ }+\text{ }{{\omega }^{2}}~=\text{ }0\text{ }and\text{ }{{\omega }^{3}}~=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-610c17f58a250f7faec93ac5ebc26a70_l3.png "Rendered by QuickLaTeX.com")
Now by simplifying we get,
Again by substituting
in above matrix we get,
Therefore
![\[LHS\text{ }=\text{ }RHS\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cced951fcde6aafcdd3af215898b2124_l3.png "Rendered by QuickLaTeX.com")
Hence the proof.
If ω is a complex cube root of unity, show that
If ω is a complex cube root of unity, show that