Question

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

• A. $2$
• B. $3$
• C. $-1$
• D. $-4$

Answer 1

The correct option is D $-4$

$\Delta =\left|\begin{array}{ccc}{e}^{-2iA}& e^{iC}& e^{iB}\\ e^{iC}& e^{-2iB}& e^{iA}\\ e^{iB}& e^{iA}& e^{-2iC}\end{array}\right|$

Let $\alpha =e^{iA},\beta =e^{iB}$ and $\gamma =e^{iC}$, then

$\alpha \beta \gamma =e^{i(A+B+C)}=e^{i\pi }=\cos \pi +i\sin \pi =-1$

$\Delta =\left|\begin{array}{ccc}\frac{1}{{\alpha }^2}& \gamma & \beta \\ \gamma & \frac{1}{{\beta }^2}& \alpha \\ \beta & \alpha & \frac{1}{{\gamma }^2}\end{array}\right|$

$=\frac{1}{{\alpha }^2{\beta }^2{\gamma }^2}\left|\begin{array}{ccc}1& {\alpha }^2\gamma & {\alpha }^2\beta \\ {\beta }^2\gamma & 1& {\beta }^2\alpha \\ {\gamma }^2\beta & {\gamma }^2\alpha & 1\end{array}\right|=\left|\begin{array}{ccc}1& -\frac{\alpha }{\beta }& -\frac{\alpha }{\gamma }\\ -\frac{\beta }{\gamma }& 1& -\frac{\beta }{y}\\ -\frac{\gamma }{\alpha }& -\frac{\gamma }{\beta }& 1\end{array}\right|$

$=\frac{1}{\alpha \beta \gamma }\left|\begin{array}{ccc}\alpha & -\alpha & -\alpha \\ -\beta & \beta & -\beta \\ -\gamma & -\gamma & \gamma \end{array}\right|=\frac{\alpha \beta \gamma }{\alpha \beta \gamma }\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|$

Applying $C_1\to C_1+C_2$

$\Delta =\left|\begin{array}{ccc}0& -1& -1\\ 0& 1& -1\\ -2& -1& 1\end{array}\right|=-2(1+1)=-4$