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

Answer 1

Compute the required value.

Given: $A,B,C$ are angles of a triangle.

Since, the sum of all interior angle of triangle is $180°$ or $\mathrm{\pi }$.

Thus,

$$\begin{gathered} \Rightarrow A+B+C=\mathrm{\pi } \\ \Rightarrow \mathrm{B}+\mathrm{C}=\mathrm{\pi }-\mathrm{A} \end{gathered}$$

Take exponential both side,

$$\begin{gathered} \Rightarrow {\mathrm{e}}^{-\mathrm{i}(\mathrm{B}+\mathrm{C})}={\mathrm{e}}^{-\mathrm{i}(\mathrm{\pi }-\mathrm{A})} \\ \Rightarrow {\mathrm{e}}^{-\mathrm{i}(\mathrm{B}+\mathrm{C})}=-{\mathrm{e}}^{\mathrm{iA}} \end{gathered}$$ ; $\left[e^{-i\mathrm{\pi }}=\cos \left(-\pi \right)-i\sin \left(-\mathrm{\pi }\right)=-1\right]$

We want to calculate the value of $∆=\left|\begin{array}{ccc}{e}^{2iA}& e^{-iC}& e^{-iB}\\ e^{-iC}& e^{2iB}& e^{-iA}\\ e^{-iB}& e^{-iA}& e^{2ic}\end{array}\right|$

Taking $e^{iA},e^{iB},e^{iC}$ common from $R_1,R_2,R_3$ respectively,

$$\begin{gathered} ∆=e^{iA}{e}^{iB}{e}^{iC}\left|\begin{array}{ccc}{e}^{iA}& e^{-i(A+C)}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{iB}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{-i(A+C)}& e^{iC}\end{array}\right| \\ ∆=e^{i(A+B+C)}\left|\begin{array}{ccc}{e}^{iA}& e^{-i(A+C)}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{iB}& e^{-i(A+B)}\\ e^{-i(B+C)}& e^{-i(A+C)}& e^{iC}\end{array}\right|\left[A+B+C=\mathrm{\pi }\right] \\ ∆=e^{i\mathrm{\pi }}\left|\begin{array}{ccc}{e}^{iA}& e^{-i(\mathrm{\pi }-B)}& e^{-i(\mathrm{\pi }-C)}\\ e^{-i(\mathrm{\pi }-\mathrm{A})}& e^{iB}& e^{-i(\mathrm{\pi }-C)}\\ e^{-i(\mathrm{\pi }-\mathrm{A})}& e^{-i(\mathrm{\pi }-B)}& e^{iC}\end{array}\right|\left[e^{i\mathrm{\pi }}=\cos \left(\mathrm{\pi }\right)+i\sin \left(\mathrm{\pi }\right)=-1\right] \\ ∆=-\left|\begin{array}{ccc}{e}^{iA}& e^{-i\mathrm{\pi }}{e}^{iB}& e^{-i\mathrm{\pi }}{e}^{iC}\\ e^{-i\mathrm{\pi }}{e}^{iA}& e^{iB}& e^{-i\mathrm{\pi }}{e}^{iC}\\ e^{-i\mathrm{\pi }}{e}^{iA}& e^{-i\mathrm{\pi }}{e}^{iB}& e^{iC}\end{array}\right| \\ ∆=-\left|\begin{array}{ccc}{e}^{iA}& -e^{iB}& -e^{iC}\\ -e^{iA}& e^{iB}& -e^{iC}\\ -e^{iA}& -e^{iB}& e^{iC}\end{array}\right| \end{gathered}$$

Taking $e^{iA},e^{iB},e^{iC}$ common from $C_1,C_2,C_3$ respectively,

$$\begin{gathered} ∆=-e^{iA}{e}^{iB}{e}^{iC}\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right| \\ ∆=-e^{i(A+B+C)}\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right|\left[A+B+C=\mathrm{\pi }\right] \\ ∆=-e^{i\mathrm{\pi }}\left|\begin{array}{ccc}1& -1& -1\\ -1& 1& -1\\ -1& -1& 1\end{array}\right| \\ ∆=1(1-1)+1(-1-1)-1(1+1) \\ ∆=0-2-2 \\ ∆=-4 \end{gathered}$$

Hence, 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 $-4$.