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Mathematics · Determinants

JEE Main 2024 — 6 April, Shift 1 — Question 8

For α,β∈R\alpha, \beta \in \mathrm{R} and a natural number n , let Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣{{A}_{r}}=\left| \begin{matrix}r & 1 & \frac{{{n}^{2}}}{2}+\alpha \\2r & 2 & {{n}^{2}}-\beta \\3r-2 & 3 & \frac{n\left( 3n-1 \right)}{2} \\\end{matrix} \right|.Then 2A10−A82 A_{10}-A_{8} is

  1. Option A:

    4α+2β4 \alpha+2 \beta

    Correct
  2. Option B:

    2α+4β2 \alpha+4 \beta

  3. Option C:

    2n2 n

  4. Option D:

    0

Answer: A

Step-by-step solution

Ar=∣r1n22+α2r2n2−β3r−23n(3n−1)2∣A_r = \begin{vmatrix} r & 1 & \dfrac{n^{2}}{2} + \alpha \\ 2r & 2 & n^{2} - \beta \\ 3r - 2 & 3 & \dfrac{n(3n - 1)}{2} \end{vmatrix} 2A10−A8=∣201n22+α402n2−β563n(3n−1)2∣−∣81n22+α162n2−β223n(3n−1)2∣2A_{10} - A_8 = \begin{vmatrix} 20 & 1 & \dfrac{n^{2}}{2} + \alpha \\ 40 & 2 & n^{2} - \beta \\ 56 & 3 & \dfrac{n(3n - 1)}{2} \end{vmatrix} - \begin{vmatrix} 8 & 1 & \dfrac{n^{2}}{2} + \alpha \\ 16 & 2 & n^{2} - \beta \\ 22 & 3 & \dfrac{n(3n - 1)}{2} \end{vmatrix} ⇒∣241n22+α242n2−β343n(3n−1)2∣\Rightarrow \begin{vmatrix} 24 & 1 & \dfrac{n^{2}}{2} + \alpha \\ 24 & 2 & n^{2} - \beta \\ 34 & 3 & \dfrac{n(3n - 1)}{2} \end{vmatrix} ⇒∣01n22+α02n2−β−23n(3n−1)2∣\Rightarrow \begin{vmatrix} 0 & 1 & \dfrac{n^{2}}{2} + \alpha \\ 0 & 2 & n^{2} - \beta \\ -2 & 3 & \dfrac{n(3n - 1)}{2} \end{vmatrix} ⇒−2[(n2−β)−(n2+2α)]\Rightarrow -2\Big[ (n^{2} - \beta) - (n^{2} + 2\alpha) \Big] ⇒−2(−β−2α)=4α+2β\Rightarrow -2(-\beta - 2\alpha) = 4\alpha + 2\beta

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Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Mathematics
Chapter
Determinants
Topic
Determinants

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