# If $A,B,C$ are angles of a triangle then $\begin{vmatrix}e^{2iA}&e^{-iC}&e^{-iB}\\e^{-iC}&e^{2iB}&e^{-iA}\\e^{-iB}&e^{-iA}&e^{2iC}\end{vmatrix}$ is equal to

$(a)\;4\qquad(b)\;0\qquad(c)\;-4\qquad(d)\;-1$

Since $A+B+C=\pi$ and $e^{i\pi}=\cos \pi+i\sin \pi=-1$
$e^{i(B+C)}=e^{i(\pi-A)}=-e^{-iA}$
$e^{-i(B+C)}=-e^{iA}$
By taking $e^{iA},e^{iB},e^{-ic}$ common from $R_1,R_2$ and $R_3$ respectively ,we have
$\Delta=-\begin{vmatrix}e^{iA}&e^{-i(A+C)}&e^{-i(A+B)}\\e^{-i(B+C)}&e^{iB}&e^{-i(A+B)}\\e^{-i(B+C)}&e^{-i(A+C)}&e^{iC}\end{vmatrix}$
$\quad=\begin{vmatrix}e^{iA}&-e^{iB}&-e^{iC}\\-e^{iA}&e^{iB}&-e^{iC}\\-e^{iA}&-e^{iB}&e^{iC}\end{vmatrix}$
By taking $e^{iA},e^{iB},e^{iC}$ common from $C_1,C_2$ and $C_3$ respectively we have
$\Delta=(+1)\begin{vmatrix}1&-1&-1\\-1&1&-1\\-1&-1&1\end{vmatrix}=-4$
Hence (c) is the correct answer.
answered Nov 25, 2013
edited Mar 20, 2014