Question

Suppose $A, B, C$ are angles of a triangle, and let
$\Delta=\begin{vmatrix} e^{2 i A} & e^{-i C} & e^{-i B} \\ e^{-i C} & e^{2 i B} & e^{-i A} \\ e^{-i B} & e^{-i A} & e^{2 i C} \end{vmatrix}$
Then value of $\Delta$ is

• A. -1
• B. -4
• C. 0
• D. 4

Answer 1

Answer: (B)

Taking $e^{i A}$, common from $R_1, e^{i B}$ from $R_2$ and $e^{i C}$ from $R_3$, we get
$\Delta=e^{i(A+B+C)} \Delta_1$
where
$\Delta_1=\begin{vmatrix}e^{i A} & e^{-i(A+C)} & e^{-i(A+B)} \\e^{-i(B+C)} & e^{i B} & e^{-i(A+B)} \\e^{-i(B+C)} & e^{-i(A+C)} & e^{i C}\end{vmatrix}$
But $A+B+C=\pi$, so that $e^{i(A+B+C)}=e^{i \pi}$
$=\cos \pi+i \sin \pi=-1 \text {. Also, }$
$A+C=\pi-B \Rightarrow e^{-i(A+C)}=e^{-\pi i} e^{i B}=-e^{i B} .$
Thus,$\Delta_1 =\begin{vmatrix}e^{i A} & -e^{i B} & -e^{i C} \\-e^{-i A} & e^{i B} & -e^{i C} \\-e^{i A} & -e^{i B} & e^{i C}\end{vmatrix}$
$ =e^{i(A+B+C)}\begin{vmatrix}1 & -1 & -1 \\-1 & 1 & -1 \\-1 & -1 & 1\end{vmatrix}$
Using $C_1 \rightarrow C_1+C_2$, we get
$\Delta_1=(-1)\begin{vmatrix}0 & -1 & -1 \\0 & 1 & -1 \\-2 & -1 & 1
\end{vmatrix}=(-1)(-2)(2)=4$
Therefore, $\Delta=(-1) \Delta_1=-4$