ExamHabit
Practise free

JEE Main 2014

JEE Main 2014 Mathematics question on Matrices and Determinants

If α,β≠0\alpha, \beta \neq 0 and f(n)=αn+βnf(n) = \alpha^n + \beta^n and ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣=K(1−α)2(1−β)2(α−β)2\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix} = K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2, then KK is equal to
  1. A.
    αβ\alpha\beta
  2. B.
    1αβ\dfrac{1}{\alpha\beta}
  3. C.
    11
  4. D.
    −1-1
Show answer and solution

Correct option: C

Solution

The determinant can be factored by recognizing it equals the Gram-style determinant of the vector (1, α, β) and (1, 1, 1). Specifically, the matrix can be written as the product:

|1 1 1 | |1 1 1 |
|1 α β | × |1 α β |
|1 α² β²| |1 α² β² |

Each entry (i,j) of this product is 1 + αⁱ⁺ʲ⁻² + βⁱ⁺ʲ⁻², matching 1+f(i+j-2), which reproduces the given matrix entries. The determinant of the matrix equals the square of the determinant of the 3×3 Vandermonde-type matrix V with rows (1,1,1), (1,α,β), (1,α²,β²). The determinant of V is the Vandermonde determinant (α-1)(β-1)(β-α), so det(matrix) = [(α-1)(β-1)(β-α)]² = (1-α)²(1-β)²(α-β)². Comparing with K(1-α)²(1-β)²(α-β)² gives K = 1.

[JEE Main 2014]

Practise Matrices and Determinants the way it is asked

A timed set of real JEE Main Matrices and Determinants questions, graded as the exam grades. No sign-up.

Start a Matrices and Determinants test

More: all JEE Main Matrices and Determinants questions · JEE Main 2014 · Mathematics chapter list