SATHEE: Matrices and Determinants 2 Question 12

Matrices and Determinants 2 Question 12

14. If $α, β \neq 0$ and $f (n) = α^{n} + β^{n}$ and

$| \begin{matrix} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{matrix} |$

$= K (1 - α)^{2} (1 - β)^{2} (α - β)^{2}$ , then $K$ is equal to

(2014 Main)

(a) $α β$

(b) $\frac{1}{α β}$

(d) -1

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Answer:

Correct Answer: 14. (c)

Solution:

PLAN: Use the property that, two determinants can be multiplied column-to-row or row-to-column, to write the given determinant as the product of two determinants and then expand.

Given, $f (n) = α^{n} + β^{n}, f (1) = α + β, f (2) = α^{2} + β^{2}$ ,

$\begin{aligned} f (3) = α^{3} + β^{3}, f (4) = α^{4} + β^{4} \\ Let & Δ = | \begin{array}{c} 3 & 1 + f (1) & 1 + f (2) \\ 1 + f (1) & 1 + f (2) & 1 + f (3) \\ 1 + f (2) & 1 + f (3) & 1 + f (4) \end{array} | \\ \Rightarrow Δ & = | \begin{array}{c} 3 & 1 + α + β & 1 + α^{2} + β^{2} \\ 1 + α + β & 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} \\ 1 + α^{2} + β^{2} & 1 + α^{3} + β^{3} & 1 + α^{4} + β^{4} \end{array} | \end{aligned}$

$= | \begin{matrix} 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 & 1 \cdot 1 + 1 \cdot α + 1 \cdot β \\ 1 \cdot 1 + 1 \cdot α + 1 \cdot β & 1 \cdot 1 + α \cdot α + β \cdot β \\ 1 \cdot 1 + 1 \cdot α^{2} + 1 \cdot β^{2} & 1 \cdot 1 + α^{2} \cdot α + β^{2} \cdot β \end{matrix} |$ $| \begin{matrix} 1 \cdot 1 + 1 \cdot α^{2} + 1 \cdot β^{2} \\ 1 \cdot 1 + α \cdot α^{2} + β \cdot β^{2} \\ 1 \cdot 1 + α^{2} \cdot α^{2} + β^{2} \cdot β^{2} \end{matrix} |$

$= | \begin{matrix} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{matrix} | | \begin{matrix} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{matrix} | = {| \begin{matrix} 1 & 1 & 1 \\ 1 & α & β \\ 1 & α^{2} & β^{2} \end{matrix} |}^{2}$