Matrices & Determinants
Properties of determinants
Grade Class 12
Question:
If α, β ≠ 0, and f(n) = α^n + β^n and <br> <img src="https://latex.codecogs.com/svg.latex?\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 K is equal to :
(1) αβ
(2) 1/αβ
(3) 1
(4) -1
Step-by-Step Solution
Key Concept: The determinant can be expressed as the product of two matrices by decomposing the entries 1+f(n) = 1 + \alpha^n + \beta^n = 1^n + \alpha^n + \beta^n. This allows the determinant to be written as the product of two Vandermonde-like matrices.
The given determinant can be written as the product of two matrices: <br> <img src="https://latex.codecogs.com/svg.latex?\begin{vmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{vmatrix} \begin{vmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{vmatrix}"> <br> The value of this determinant is (1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2. Comparing this with the given expression K(1-\alpha)^2 (1-\beta)^2 (\alpha-\beta)^2, we get K = 1.
Correct Answer: 3