Question

If $A,B,C$ are the angles of a triangle, and
$\Delta =\left|\begin{array}{ccc}{e}^{-2iA}& e^{iC}& e^{iB}\\ e^{iC}& e^{-2iB}& e^{iA}\\ e^{iB}& e^{iA}& e^{-2iC}\end{array}\right|$
find $\left|\Delta \right|$

Answer 1

$A,B,Caretheanglesofatriangle,and\Delta =\left|\begin{array}{ccc}{e}^{-2iA}& e^{iC}& e^{iB}\\ e^{iC}& e^{-2iB}& e^{iA}\\ e^{iB}& e^{iA}& e^{-2iC}\end{array}\right|Let\alpha =e^{iA},\beta =e^{iB}and\gamma =e^{iC}then\alpha \beta \gamma =e^{i(A+B+C)}=e^{i\pi }=\cos \pi +i\sin \pi =-1and\Delta =\left|\begin{array}{ccc}1/{\alpha }^2& \gamma & \beta \\ \gamma & 1/{\beta }^2& \alpha \\ \beta & \alpha & 1/{\gamma }^2\end{array}\right|=\frac{1}{{\alpha }^2{\beta }^2{\gamma }^2}\left|\begin{array}{ccc}1& {\alpha }^2\gamma & {\alpha }^2\beta \\ {\beta }^2\gamma & 1& {\beta }^2\alpha \\ {\gamma }^2\beta & {\gamma }^2\alpha & 1\end{array}\right|=\left|\begin{array}{ccc}1& -\alpha /\beta & -\alpha /\gamma \\ -\beta /\alpha & 1& -\beta /\gamma \\ -\gamma /\alpha & -\gamma /\beta & 1\end{array}\right|=\frac{1}{\alpha \beta \gamma }\left|\begin{array}{ccc}\alpha & -\alpha & -\alpha \\ -\beta & \beta & -\beta \\ -\gamma & -\gamma & \gamma \end{array}\right|\left[multiplying{C}_1\left\{C_2\right\}\left[C_3\right]by\alpha \left\{\beta \right\}\left[\gamma \right]\right]=\frac{\alpha \beta \gamma }{\alpha \beta \gamma }\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|=\left|\begin{array}{ccc}0& -1& -1\\ 0& 1& -1\\ -2& -1& 1\end{array}\right|=-2(1+1)=-4\therefore \left|\Delta \right|=4$