If z - e2iA e-iC e-iBe-iC e2iB e-iAe-iB e-iA e2iC- where A-B-C - - and ei- - cos- -i sin- then

Taking e^iA, e^iB, e^iC out of the determinant, z = e^iA· e^iB· e^iC e^iA amp; e^ i(C+A) amp; e^ i(B+A) e^ i(C+B) amp; e^iB amp; e^ i(A+B) e^ i(B+C) ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

If z = e2iA ...

Question

If $z = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{2 i A} & e^{- i C} & e^{- i B} e^{- i C} & e^{2 i B} & e^{- i A} e^{- i B} & e^{- i A} & e^{2 i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$ , where $A + B + C = π$ and $e^{i θ} = cos θ + i sin θ$ then

R e (z) = 4

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R e (z) = - 4

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I m (z) = 1

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I m (z) = - 1

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Solution

The correct option is B $R e (z) = - 4$
Taking $e^{i A}, e^{i B}, e^{i C}$ out of the determinant,
$z = e^{i A} \cdot e^{i B} \cdot e^{i C} ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i A} & e^{- i (C + A)} & e^{- i (B + A)} e^{- i (C + B)} & e^{i B} & e^{- i (A + B)} e^{- i (B + C)} & e^{- i (A + C)} & e^{i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$
We know that
$A + B + C = π$
So,
$\Rightarrow z = e^{i (A + B + C)} ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i A} & e^{- i (π - B)} & e^{- i (π - C)} e^{- i (π - A)} & e^{i B} & e^{- i (π - C)} e^{- i (π - A)} & e^{- i (π - B)} & e^{i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Now,
$e^{- i (π - A)} = cos (A - π) + i sin (A - π) = - cos A - i sin A = - e^{i A}$
$e^{i (A + B + C)} = e^{i π} = - 1$
$\Rightarrow z = (- 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i A} & - e^{i B} & - e^{i C} - e^{i A} & e^{i B} & - e^{i C} - e^{i A} & - e^{i B} & e^{i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Using row operations,
$R_{3} \to R_{3} + R_{2}$
$\Rightarrow Z = (- 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{i A} & - e^{i B} & - e^{i C} - e^{i A} & e^{i B} & - e^{i C} - 2 e^{i A} & 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$\Rightarrow z = (- 1) [- 2 e^{i A} (2 e^{i (B + C)})] \Rightarrow z = 4 e^{i (A + B + C)} = - 4$

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If z=∣∣ ∣ ∣∣e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iAe2iC∣∣ ∣ ∣∣, where A+B+C=π and eiθ=cosθ+isinθ then

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