Suppose A, B, C are angles of a triangle, and let Δ=e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iAe2iCwhere eiθ=cosθ+isinθ and i2=−1Then value of Δ is

Taking eiA , common from R1,e fromR2 and from , we get Δ=eiA+B+CΔ1whereΔ1=eiAe−iA+Ce−iA+Be−iB+CeiBe−iA+Be−iB+Ce−iA+CeiCBut A+ B+ C =π, so that eiA+B+C=eiπ=cosπ+isinπ=−1.Also, A+C = π−B ⇒e−i(A+C)=e−πieiB=−eiBThus, Δ1=eiA−eiB−eiC−eiAeiB−eiC−eiA−eiBeiC=eiA+B+C1−1−1−11−1−1−11Using C1→C1+C2, we get Δ1=−10−1−101−1−2−11=−1−22=4Therefore , Δ=−1Δ1=−4

Suppose A, B, C are angles of a triangle, and let Δ=e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iAe2iCwhere eiθ=cosθ+isinθ and i2=−1Then value of Δ is

Taking eiA , common from R1,e fromR2 and from , we get Δ=eiA+B+CΔ1whereΔ1=eiAe−iA+Ce−iA+Be−iB+CeiBe−iA+Be−iB+Ce−iA+CeiCBut A+ B+ C =π, so that eiA+B+C=eiπ=cosπ+isinπ=−1.Also, A+C = π−B ⇒e−i(A+C)=e−πieiB=−eiBThus, Δ1=eiA−eiB−eiC−eiAeiB−eiC−eiA−eiBeiC=eiA+B+C1−1−1−11−1−1−11Using C1→C1+C2, we get Δ1=−10−1−101−1−2−11=−1−22=4Therefore , Δ=−1Δ1=−4

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Suppose A, B, C are angles of a triangle, and let $Δ = |\begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} \\ e^{- i C} & e^{2 i B} & e^{- i A} \\ e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix}|$
where $e^{i θ} = \cos θ + i \sin θ {and i}^{2} = - 1$
Then value of $Δ$ is

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-4

answer is B.

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Detailed Solution

Taking $e^{i A}$ , common from $R_{1}, e$ from $R_{2}$ and from , we get $Δ = e^{i (A + B + C)} Δ_{1}$
where
$Δ_{1} = |\begin{matrix} e^{i A} & e^{- i (A + C)} & e^{- i (A + B)} \\ e^{- i (B + C)} & e^{i B} & e^{- i (A + B)} \\ e^{- i (B + C)} & e^{- i (A + C)} & e^{i C} \end{matrix}|$
But A+ B+ C = $π,$ so that $e^{i (A + B + C)} = e^{i π}$ $= \cos π + i \sin π = - 1.$
Also, A+C = $π - B \Rightarrow e^{- i (A + C)} = e^{- π i} e^{i B} = - e^{i B}$
Thus, $Δ_{1} = |\begin{matrix} e^{i A} & - e^{i B} & - e^{i C} \\ - e^{i A} & e^{i B} & - e^{i C} \\ - e^{i A} & - e^{i B} & e^{i C} \end{matrix}|$ $= e^{i (A + B + C)} |\begin{matrix} 1 & - 1 & - 1 \\ - 1 & 1 & - 1 \\ - 1 & - 1 & 1 \end{matrix}|$

Using $C_{1} \to C_{1} + C_{2},$ we get $Δ_{1} = (- 1) |\begin{matrix} 0 & - 1 & - 1 \\ 0 & 1 & - 1 \\ - 2 & - 1 & 1 \end{matrix}| = (- 1) (- 2) (2) = 4$

Therefore , $Δ = (- 1) Δ_{1} = - 4$