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Properties of Determinants

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Question

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of determinant $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s (α - β) & c o s α c o s (α - β) & 1 & c o s β c o s α & c o s β & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is equal to

A

a + b

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B

0

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C

a – b

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D

a + b + c

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Solution

The correct option is B
0

We have $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & c o s (α - β) & c o s α c o s (α - β) & 1 & c o s β c o s α & c o s β & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ [E x p a n d i n g a l o n g R_{1}]$
$= (1 - c o s^{2} β) + c o s (α - β) [(c o s α) . (c o s β) - c o s (α - β) - c o s α c o s β] = s i n^{2} β + c o s (α - β) [2 c o s α c o s β - c o s (α - β)] = s i n^{2} β + c o s (α - β) c o s (α + β) - c o s^{2} α = s i n^{2} β + (c o s^{2} α - s i n^{2} β) - c o s^{2} α = 0$

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Similar questions

f (α) = ⎡ ⎢ ⎣ \begin{matrix} c o s α & - s i n α & 0 s i n α & c o s α & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦

(f (α))^{- 1}

$A (α, β) = ⎛ ⎜ ⎝ \begin{matrix} c o s α & s i n α & 0 - s i n α & c o s α & 0 0 & 0 & e^{β} \end{matrix} ⎞ ⎟ ⎠ \Rightarrow {[A (α, β)]}^{- 1} =$

α, β

are non real numbers satisfying

x^{3} - 1 = 0

then the value of

∣ ∣ ∣ ∣ \begin{matrix} λ + 1 & α & β α & λ + β & 1 β & 1 & λ + α \end{matrix} ∣ ∣ ∣ ∣

Q. Question 16

a^{2} - b^{2}

(a - b)^{2}

(a - b) (a - b)

(a + b) (a - b)

(a + b) (a + b)

Using properties of determinants prove the following questions.

$∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n α & c o s α & c o s (α + δ) s i n β & c o s β & c o s (β + δ) s i n γ & c o s γ & c o s (γ + δ) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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