First slide

Introduction to Determinants

Question

If $a = \cos θ + isin θ, b = \cos 2 θ - isin 2 θ, c = \cos 3 θ$ $+ isin 3 θ and if |\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}| = 0$ , then

Moderate

A

$θ = 2 kπ, k \in Z$
B

$θ = (2 k + 1) π, k \in Z$
C

$θ = (4 k + 1) π, k \in Z$
D

none of these

Solution

$\begin{array}{l} Δ = |\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}| \\ = - (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ca) \\ = - \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] = 0 \\ \Rightarrow a + b + c = 0 or a = b = c \end{array}$
$If a + b + c = 0, we have$
$\cos θ + \cos 2 θ + \cos 3 θ = 0 and \sin θ - \sin 2 θ + \sin 3 θ = 0$
or $or \cos 2 θ (2 \cos θ + 1) = 0 and \sin 2 θ (1 - 2 \cos θ) = 0$
which is not possible as $\cos 2 θ = 0$ gives $\sin 2 θ \neq 0, \cos θ \neq 1 / 2$ . And $\cos θ = - 1 / 2$ gives $\sin 2 θ \neq 0, \cos θ \neq 1 / 2 .$ . Therefore, Eq. (1) does not hold simultaneously. Therefore,
$\begin{array}{l} a + b + c \neq 0 \\ \Rightarrow a = b = c \\ ∴ e^{iθ} = e^{- 2 iθ} = e^{3 iθ} \end{array}$
which is satisfied only by $e^{iθ} = 1, i.e., \cos θ = 1, \sin θ = 0$ so $θ = 2 kπ, k \in Z$ .

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