Question

Suppose $A,B,C$ are angles of a triangle, and let
$\Delta =\left|\begin{array}{ccc}{e}^{2iA}& e^{-iC}& e^{-iB}\\ e^{-iC}& e^{2iB}& e^{-iA}\\ e^{-iB}& e^{-iA}& e^{2iC}\end{array}\right|$
Then value of $\Delta$ is

• A. -1
• B. -4
• C. 0
• D. 4

Answer 1

Answer: (B)

Taking $e^{iA}$, common from $R_1,e^{iB}$ from $R_2$ and $e^{iC}$ from $R_3$, we get
$\Delta =e^{i(A+B+C)}{\Delta }_1$
where
${\Delta }_1=\left|\begin{array}{ccc}{e}^{iA}& e^{-i(A+C)}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{iB}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{-i(A+C)}& e^{iC}\end{array}\right|$
But $A+B+C=\pi$, so that $e^{i(A+B+C)}=e^{i\pi }$
$=\cos \pi +i\sin \pi =-1\text{. Also, }$
$A+C=\pi -B\Rightarrow e^{-i(A+C)}=e^{-\pi i}{e}^{iB}=-e^{iB}.$
Thus,${\Delta }_1=\left|\begin{array}{ccc}{e}^{iA}& -e^{iB}& -e^{iC}\\ -e^{-iA}& e^{iB}& -e^{iC}\\ -e^{iA}& -e^{iB}& e^{iC}\end{array}\right|$
$=e^{i(A+B+C)}\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|$
Using $C_1\to C_1+C_2$, we get
${\Delta }_1=(-1)\left|\begin{array}{ccc}0& -1& -1\\ 0& 1& -1\\ -2& -1& 1<br/>\end{array}\right|=(-1)(-2)(2)=4$
Therefore, $\Delta =(-1){\Delta }_1=-4$