Question

If $A,B,C$ are the angles of a triangle, then the absolute value of $\left|\begin{array}{ccc}{e}^{-2iA}& e^{iC}& e^{iB}\\ e^{iC}& e^{-2iB}& e^{iA}\\ e^{iB}& e^{iA}& e^{-2iC}\end{array}\right|$ is

Answer 1

Let $\alpha =e^{iA},$ $\beta =e^{iB}$ and $\gamma =e^{iC}$
Then, $\alpha \beta \gamma =e^{i(A+B+C)}=e^{i\pi }$
$\alpha \beta \gamma =\cos \pi +i\sin \pi =-1\cdots \left(1\right)$
$\Rightarrow \Delta =\left|\begin{array}{ccc}\frac{1}{{\alpha }^2}& \gamma & \beta \\ \gamma & \frac{1}{{\beta }^2}& \alpha \\ \beta & \alpha & \frac{1}{{\gamma }^2}\end{array}\right|$
Multiplying $R_1$ by ${\alpha }^2,$ $R_2$ by ${\beta }^2$ and $R_3$ by ${\gamma }^2,$ we get
$\Delta =\frac{1}{{\alpha }^2{\beta }^2{\gamma }^2}\left|\begin{array}{ccc}1& {\alpha }^2\gamma & {\alpha }^2\beta \\ {\beta }^2\gamma & 1& {\beta }^2\alpha \\ {\gamma }^2\beta & {\gamma }^2\alpha & 1\end{array}\right|$

$\Rightarrow \Delta =\left|\begin{array}{ccc}1& -\frac{\alpha }{\beta }& -\frac{\alpha }{\gamma }\\ -\frac{\beta }{\alpha }& 1& -\frac{\beta }{\gamma }\\ -\frac{\gamma }{\alpha }& -\frac{\gamma }{\beta }& 1\end{array}\right|\left[\text{Using}\right(1\left)\right]$

Multiplying $C_1,C_2,C_3$ by $\alpha \beta ,\gamma$ respectively,
$\Delta =\frac{1}{\alpha \beta \gamma }\left|\begin{array}{ccc}\alpha & -\alpha & -\alpha \\ -\beta & \beta & -\beta \\ -\gamma & -\gamma & \gamma \end{array}\right|$
Taking $\alpha ,\beta ,\gamma$ from $R_1,R_2,R_3$ respectively,
$\Delta =\frac{\alpha \beta \gamma }{\alpha \beta \gamma }\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|=-4$
$\therefore |\Delta |=4$