Suppose A, B, C are angles of a triangle, and let Δ=e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iAe2iC then value of ∆ is

Talking eiA, common from R1,eiBfrom R2 and eiCfrom R3 we getΔ=ei(A+B+C)Δ1where Δ1=eiAei(A+C)ei(A+B)e−i(B+C)eiBe−i(A+B)e−i(B+C)e−i(A+C)eiCBut A + B + C =π, so that ei(A+R+C)=eiπ=cos⁡π+isin⁡π=−1.AlsoA+C=π−B ⇒ e−i(A+C)=e−πieiB=−eiBThus, Δ1=eiA−eiB−eiC−e−iAeiB−eiC−eiA−eiBeiC=ei(A+B+C)1−1−1−11−1−1−11Using C1→C1+C2 we getΔ1=(−1)0−1−101−1−2−11=(−1)(−2)(2)=4Therefor Δ=(−1)Δ1=−4