Lf - is a cube root of unity- then -x b c b c -x c -x b is equal to-

The correct option is B (x b c)(x bω cω^2)( x bω^2 cω) ω is a cube root of unity. Δ= x amp; b amp; c b amp; c amp; x c ...

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Standard XII

Mathematics

Special Determinants

lf ω is a c...

Question

lf $ω$ is a cube root of unity, then $∣ ∣ ∣ ∣ \begin{matrix} - x & b & c b & c & - x c & - x & b \end{matrix} ∣ ∣ ∣ ∣$ is equal to:

(x - b - c) (x - b ω - c ω^{2}) (x + b + c)

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(x - b - c) (x - b ω - c ω^{2}) (x - b ω^{2} - c ω)

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(x - b - c) (x - b ω^{2} - c ω) (x + b + c)

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Solution

The correct option is B $(x - b - c) (x - b ω - c ω^{2}) (x - b ω^{2} - c ω)$
$ω$ is a cube root of unity.
$Δ = ∣ ∣ ∣ ∣ \begin{matrix} - x & b & c b & c & - x c & - x & b \end{matrix} ∣ ∣ ∣ ∣$
Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ and expanding gives
$Δ = (x - b - c) [(x + b) (x + c) + {(b - c)}^{2}]$
$(x - b ω - c ω^{2}) (x - b ω^{2} - c ω) = x^{2} - b x (ω + ω^{2}) - c x (ω + ω^{2}) + b c (ω + ω^{2}) + b^{2} ω^{3} + c^{2} ω^{3} = (x + b) (x + c) + {(b - c)}^{2} . . . (∵ 1 + ω + ω^{2} = 0 a n d ω^{3} = 1)$
$∴ Δ = (x - b - c) (x - b ω - c ω^{2}) (x - b ω^{2} - c ω)$

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Similar questions

If $ω$ is a complex cube root of unity, then the value of $\frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}}$ + $\frac{a + b ω + c ω^{2}}{b + c ω + b ω^{2}}$ is :

Q. If cube root of unity is

1, ω, ω^{2}

then find the value of:

\frac{(a + b ω + c ω^{2})}{(c + a ω + b ω^{2})} + \frac{(a + b ω + c ω^{2})}{(b + c ω + a ω^{2})}

Q. If

p = a + b ω + c ω^{2}, q = b + c ω + a ω^{2}

, and

r = c + a ω + b ω^{2}

, where

a, b, c \neq 0

and

ω

is the complex cube root of unity, then

Q. If

p = a + b ω + c ω^{2}; q = b + c ω + a ω^{2} a n d r = c + a ω + b ω^{2}

, where

a, b, c \neq 0

and

ω

is the non-real complex cube root of unity, then:

Q. If

ω

is a cube root of unity, and the value of

(1 - ω) (1 - ω^{2}) (1 - ω^{4}) (1 - ω^{8}) = α,

and

\frac{a + b ω + c ω^{2}}{b + c ω + a ω^{2}} + \frac{a + b ω + c ω^{2}}{c + a ω + b ω^{2}} = β

then

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lf ω is a cube root of unity, then ∣∣ ∣∣−xbcbc−xc−xb∣∣ ∣∣ is equal to:

lf $ω$ is a cube root of unity, then $∣ ∣ ∣ ∣ \begin{matrix} - x & b & c b & c & - x c & - x & b \end{matrix} ∣ ∣ ∣ ∣$ is equal to: