Question

If $A,B,C$ are angles of a triangle, then the 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

• A. $1$
• B. $-1$
• C. $-2$
• D. $-4$

Answer 1

The correct option is D $-4$

since $A+B+C=\pi$ and $e^{i\pi }=\cos \pi +i\sin \pi =-1$

$e^{-i(B+C)}=e^{-i(\pi -A)}=-e^{iA}$ and $e^{-i(B+C)}=-e^{iA}$

By taking $e^{iA},e^{iB},e^{iC}$ common from $R_1,R_2$ and $R_3$ respectively, we have

$\Delta =-\left|\begin{array}{ccc}{e}^{iA}& e^{-i(A+C)}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{iB}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{-iA(A+C)}& e^{iC}\end{array}\right|=-\left|\begin{array}{ccc}{e}^{iA}& -e^{iB}& -e^{iC}\\ -e^{iA}& e^{iB}& -e^{iC}\\ -e^{iA}& -e^{iB}& e^{iC}\end{array}\right|$

by taking $e^{iA},e^{iB},e^{iC}$ common from $C_1,C_2$ and $C_3$ respectively

we have

$\Delta =\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|=-4$