Let - - Ax x2 1 By y2 1 Cz z2 1 and -1 - A B C x y z zy zx xy then

The correct option is D Δ_1 = Δ = Ax amp; x ^ 2 amp; 1 By amp; y ^ 2 amp; 1 Cz amp; z ^ 2 amp; 1 #160; _ 1 = A ...

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Properties of Determinants

Let Δ = Ax ...

Question

Let $Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} A x & x^{2} & 1 B y & y^{2} & 1 C z & z^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ and $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} A & B & C x & y & z z y & z x & x y \end{matrix} ∣ ∣ ∣ ∣$ then

Δ_{1} = Δ

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Δ_{1} = 2 Δ

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Δ_{1} = - Δ

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Δ_{1} \neq Δ

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Solution

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Similar questions

If ∆ = |\begin{matrix} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{matrix}|, ∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{matrix}|, then prove that ∆ + ∆_{1} = 0 .

Δ =

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

Δ_{1}

∣ ∣ ∣ ∣ \begin{matrix} a_{1} + p b_{1} & b_{1} + q c_{1} & c_{1} + r a_{1} a_{2} + p b_{2} & b_{2} + q c_{2} & c_{2} + r a_{2} a_{3} + p b_{3} & b_{3} + q c_{3} & c_{3} + r a_{3} \end{matrix} ∣ ∣ ∣ ∣

Δ_{1} =

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} A x & x^{2} & 1 B y & y^{2} & 1 C z & z^{2} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} A & B & C x & y & z z y & z x & x y \end{matrix} ∣ ∣ ∣ ∣

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{2} x & x^{2} & 1 x^{2} & 1 & x \end{matrix} ∣ ∣ ∣ ∣ ∣ = - 7

Δ_{1} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{3} - 1 & 0 & x - x^{4} 0 & x - x^{4} & x^{1} - 1 x - x^{4} & x^{3} - 1 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

a .

360 - 2 (α + β)

b .

α + β = θ

c .

δ = δ_{1} + δ_{2}

d .

δ = 360 - 2 θ

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