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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Determinants

NDA 2017 · Mathematics · Determinants

The value of the determinant | array ccc 1-α & α-α^2 & α^2 1-β &...

The value of the determinant \left|1−αα−α2α21−ββ−β2β21−γγ−γ2γ2\begin{array}{ccc}1-\alpha & \alpha-\alpha^2 & \alpha^2 \\ 1-\beta & \beta-\beta^2 & \beta^2 \\ 1-\gamma & \gamma-\gamma^2 & \gamma^2\end{array}\right| is equal to

Options

  1. A(α−β)(β−γ)(α−γ)(\alpha-\beta)(\beta-\gamma)(\alpha-\gamma)
  2. B(α−β)(β−γ)(γ−α)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)
  3. C(α−β)(β−γ)(γ−α)(α+β+γ)\quad(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)
  4. D00

Correct answer

B. (α−β)(β−γ)(γ−α)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)

Step-by-step solution

∣1−αα−α2α21−ββ−β2β21−γγ−γ2γ2∣c1→c1+c2+c3=∣1α−α2α21β−β2β21γ−γ2γ2∣\begin{aligned} & \left|\begin{array}{ccc}1-\alpha & \alpha-\alpha^2 & \alpha^2 \\ 1-\beta & \beta-\beta^2 & \beta^2 \\ 1-\gamma & \gamma-\gamma^2 & \gamma^2\end{array}\right| \\ & c_1 \rightarrow c_1+c_2+c_3 \\ & =\left|\begin{array}{ccc}1 & \alpha-\alpha^2 & \alpha^2 \\ 1 & \beta-\beta^2 & \beta^2 \\ 1 & \gamma-\gamma^2 & \gamma^2\end{array}\right|\end{aligned}c2→c2+c3=∣1αα21ββ21γγ2∣=(α−β)(β−γ)(γ−α)\begin{aligned} & c_2 \rightarrow c_2+c_3 \\ & =\left|\begin{array}{lll}1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2 \\ 1 & \gamma & \gamma^2\end{array}\right|=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\end{aligned}
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