Question

If $\omega$ is one of the imaginary cube roots of unity, show that the square of
$\left|\begin{array}{cccc}1& \omega & {\omega }^2& {\omega }^3\\ \omega & {\omega }^2& {\omega }^3& 1\\ {\omega }^2& {\omega }^3& 1& \omega \\ {\omega }^3& 1& \omega & {\omega }^2\end{array}\right|=\left|\begin{array}{cccc}1& 1& -2& 1\\ 1& 1& 1& -2\\ -2& 1& 1& 1\\ 1& -2& 1& 1\end{array}\right|$;
hence show that the value of the determinant on the left is $3\sqrt{-3}$.

Answer 1

$LHS=\left|\begin{array}{cccc}1& \omega & {\omega }^2& {\omega }^3\\ \omega & {\omega }^2& {\omega }^3& 1\\ {\omega }^2& {\omega }^3& 1& \omega \\ {\omega }^3& 1& \omega & {\omega }^2\end{array}\right|\times \left|\begin{array}{cccc}1& \omega & {\omega }^2& {\omega }^3\\ \omega & {\omega }^2& {\omega }^3& 1\\ {\omega }^2& {\omega }^3& 1& \omega \\ {\omega }^3& 1& \omega & {\omega }^2\end{array}\right|$

$=\left|\begin{array}{cccc}1& \omega & {\omega }^2& 1\\ \omega & {\omega }^2& 1& 1\\ {\omega }^2& 1& 1& \omega \\ 1& 1& \omega & {\omega }^2\end{array}\right|\times \left|\begin{array}{cccc}1& \omega & {\omega }^2& 1\\ \omega & {\omega }^2& 1& 1\\ {\omega }^2& 1& 1& \omega \\ 1& 1& \omega & {\omega }^2\end{array}\right|$

$=\left|\begin{array}{cccc}1+{\omega }^2+{\omega }^4+1& \omega +{\omega }^3+{\omega }^2+1& {\omega }^2+\omega +{\omega }^2+\omega & 1+\omega +{\omega }^3+{\omega }^4\\ \omega +{\omega }^3+{\omega }^2+1& {\omega }^2+{\omega }^4+1+1& {\omega }^3+{\omega }^2+1+\omega & \omega +{\omega }^2+\omega +{\omega }^2\\ {\omega }^2+\omega +{\omega }^2+\omega & {\omega }^3+{\omega }^2+1+\omega & {\omega }^4+1+1+{\omega }^2& {\omega }^2+1+1+\omega \\ 1+\omega +{\omega }^3+{\omega }^2& \omega +{\omega }^2+\omega +{\omega }^2& {\omega }^2+1+\omega +{\omega }^3& 1+1+{\omega }^2+{\omega }^4\end{array}\right|$

$=\left|\begin{array}{cccc}1& 1& -2& 1\\ 1& 1& 1& -2\\ -2& 1& 1& 1\\ 1& -2& 1& 1\end{array}\right|=RHS$

$=\left|\begin{array}{cccc}1& 1& -2& 1\\ 1& 1& 1& -2\\ -2& 1& 1& 1\\ 1& -2& 1& 1\end{array}\right|=RHS$

$\left|\begin{array}{cccc}1& 1& -2& 1\\ 1& 1& 1& -2\\ -2& 1& 1& 1\\ 1& -2& 1& 1\end{array}\right|$

$=\left|\begin{array}{cccc}1& 1& -2& 1\\ 0& 0& 3& -3\\ 0& 3& -3& 3\\ 0& -3& 3& 0\end{array}\right|[C_2\to C_2-C_1;C_3\to C_3+2C_1;C_4\to C_4-C_1]$

$=\left|\begin{array}{ccc}0& 3& -3\\ 3& -3& 3\\ -3& 3& 0\end{array}\right|$

$=\left|\begin{array}{ccc}0& 3& 0\\ 3& -3& 0\\ -3& 3& 3\end{array}\right|[C_3\to C_3+C_2]$

$=3(9-0)=27$

$⟹\left|\begin{array}{cccc}1& \omega & {\omega }^2& {\omega }^3\\ \omega & {\omega }^2& {\omega }^3& 1\\ {\omega }^2& {\omega }^3& 1& \omega \\ {\omega }^3& 1& \omega & {\omega }^2\end{array}\right|=3\sqrt{3}$