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  4. If α,β ≠ 0 and f(n)=αn+βn and | 3&1& f(1)& 1 &f(2) [0.3em] 1 f(1)&1 &f(2)&1&f(3) [0.3em] 1&f(2)&1&f(3)&1&f(4) | =k(1-α)2(1-β)2(α-β)2 , then k is equal to

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Q. If $\alpha,\beta \neq 0\,$ and $\,f(n)=\alpha^n+\beta^n$ and $\begin{vmatrix} & 3&1& f(1)& 1 &f(2) \\[0.3em] 1 & f(1)&1 &f(2)&1&f(3)\\[0.3em] 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

JEE MainJEE Main 2014Determinants

Solution:

Given, $f(n)=\alpha^{n}+\beta^{n}, f(1)=\alpha+\beta, f(2)=\alpha^{2}+\beta^{2}$, $f(3)=\alpha^{3}+\beta^{3}, f(4)=\alpha^{4}+\beta^{4}$
Let $\Delta=\begin{vmatrix}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{vmatrix}$
$\Rightarrow \Delta=\begin{vmatrix}3 & 1+\alpha+\beta &1+\alpha^{2}+\beta^{2} \\1+\alpha+\beta & 1+\alpha^{2}+\beta^{2} & 1+\alpha^{3}+\beta^{3} \\1+\alpha^{2}+\beta^{2} & 1+\alpha^{3}+\beta^{3} & 1+\alpha^{4}+\beta^{4}\end{vmatrix}$
$\begin{vmatrix}1\cdot1+1\cdot1+1\cdot1&1\cdot1+1\cdot\alpha+1\cdot\beta&1\cdot1+1\cdot\alpha^{2}+1\cdot\beta^{2}\\ 1\cdot1+\cdot\alpha+1\cdot\beta&1\cdot1+\alpha\cdot\alpha+\beta\cdot\beta&1\cdot1+\alpha\cdot\alpha^{2}+\beta\cdot\beta^{2}\\ 1\cdot1+1\cdot\alpha^{2} + 1\cdot\beta^{2}&1\cdot1+\alpha^{2}\cdot\alpha+\beta^{2}\cdot\beta&1\cdot1+\alpha^{2}\cdot\alpha^{2}+\beta^{2}\cdot\beta^{2}\end{vmatrix}$
$=\begin{vmatrix}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^{2} & \beta^{2}\end{vmatrix}\begin{vmatrix}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^{2} & \beta^{2}\end{vmatrix}=\begin{vmatrix}1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^{2} & \beta^{2}\end{vmatrix}^{2}$
On expanding, we get $\Delta=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$
But given, $ \Delta=K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$
Hence, $K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$ $\therefore$
$K=1$