Questions
Suppose are angles of a triangle, and let then value of is
detailed solution
Correct option is B
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=cosπ+isinπ=−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=−4Similar Questions
If ,then is equal to
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