JEE Challenger
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MathMedAdv 2015 • Paper 1

Values of Parameter Alpha Satisfying Determinant Equation

Which among the given values of α\alpha satisfy the following determinant relation?

∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣=−648α\begin{vmatrix} (1 + \alpha)^2 & (1 + 2\alpha)^2 & (1 + 3\alpha)^2 \\ (2 + \alpha)^2 & (2 + 2\alpha)^2 & (2 + 3\alpha)^2 \\ (3 + \alpha)^2 & (3 + 2\alpha)^2 & (3 + 3\alpha)^2 \end{vmatrix} = -648\alpha

Options

A

−4-4

B

99

Correct
C

−9-9

Correct
D

44

Topics & Concepts

DeterminantsDeterminants

Step-by-Step Solution

To find the values of α\alpha that satisfy the given determinant equation, we consider the determinant:

Δ=∣(1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2∣\Delta = \begin{vmatrix} (1 + \alpha)^2 & (1 + 2\alpha)^2 & (1 + 3\alpha)^2 \\ (2 + \alpha)^2 & (2 + 2\alpha)^2 & (2 + 3\alpha)^2 \\ (3 + \alpha)^2 & (3 + 2\alpha)^2 & (3 + 3\alpha)^2 \end{vmatrix}

The general (i,j)(i, j)-th entry of the matrix can be expanded as:

aij=(i+jα)2=i2+2ijα+j2α2a_{ij} = (i + j\alpha)^2 = i^2 + 2ij\alpha + j^2\alpha^2

This can be expressed as the inner product of two vectors:

aij=(i22i1)(1jαj2α2)a_{ij} = \begin{pmatrix} i^2 & 2i & 1 \end{pmatrix} \begin{pmatrix} 1 \\ j\alpha \\ j^2\alpha^2 \end{pmatrix}

Therefore, the matrix can be written as the product of two matrices:

((1+α)2(1+2α)2(1+3α)2(2+α)2(2+2α)2(2+3α)2(3+α)2(3+2α)2(3+3α)2)=(122(1)1222(2)1322(3)1)(111α2α3αα24α29α2)\begin{pmatrix} (1 + \alpha)^2 & (1 + 2\alpha)^2 & (1 + 3\alpha)^2 \\ (2 + \alpha)^2 & (2 + 2\alpha)^2 & (2 + 3\alpha)^2 \\ (3 + \alpha)^2 & (3 + 2\alpha)^2 & (3 + 3\alpha)^2 \end{pmatrix} = \begin{pmatrix} 1^2 & 2(1) & 1 \\ 2^2 & 2(2) & 1 \\ 3^2 & 2(3) & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3\alpha \\ \alpha^2 & 4\alpha^2 & 9\alpha^2 \end{pmatrix}

Using the determinant property det⁡(AB)=det⁡(A)det⁡(B)\det(AB) = \det(A)\det(B), we have:

Δ=∣121441961∣×∣111α2α3αα24α29α2∣\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 4 & 4 & 1 \\ 9 & 6 & 1 \end{vmatrix} \times \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3\alpha \\ \alpha^2 & 4\alpha^2 & 9\alpha^2 \end{vmatrix}

Evaluating the first determinant:

∣121441961∣=1(4−6)−2(4−9)+1(24−36)=−2+10−12=−4\begin{vmatrix} 1 & 2 & 1 \\ 4 & 4 & 1 \\ 9 & 6 & 1 \end{vmatrix} = 1(4 - 6) - 2(4 - 9) + 1(24 - 36) = -2 + 10 - 12 = -4

Evaluating the second determinant by factoring out α\alpha from the second row and α2\alpha^2 from the third row:

∣111α2α3αα24α29α2∣=α⋅α2∣111123122232∣\begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3\alpha \\ \alpha^2 & 4\alpha^2 & 9\alpha^2 \end{vmatrix} = \alpha \cdot \alpha^2 \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1^2 & 2^2 & 3^2 \end{vmatrix}

The determinant is a standard Vandermonde determinant:

∣111123149∣=(2−1)(3−1)(3−2)=1×2×1=2\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{vmatrix} = (2 - 1)(3 - 1)(3 - 2) = 1 \times 2 \times 1 = 2

Thus, the second determinant is:

2α32\alpha^3

Multiplying the two determinants together, we get:

Δ=(−4)×(2α3)=−8α3\Delta = (-4) \times (2\alpha^3) = -8\alpha^3

According to the given equation:

Δ=−648α\Delta = -648\alpha −8α3=−648α-8\alpha^3 = -648\alpha α3−81α=0\alpha^3 - 81\alpha = 0 α(α2−81)=0\alpha(\alpha^2 - 81) = 0 α(α−9)(α+9)=0\alpha(\alpha - 9)(\alpha + 9) = 0

The roots of the equation are α=0,9,−9\alpha = 0, 9, -9.

Among the given options, the values of α\alpha are:

α=9andα=−9\alpha = 9 \quad \text{and} \quad \alpha = -9

Therefore, the correct options are B and C.

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