Matrices and Determinants 2 Question 12
14. If $\alpha, \beta \neq 0$ and $f(n)=\alpha^{n}+\beta^{n}$ and
$\begin{vmatrix}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{vmatrix}$
$=K(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$, then $K$ is equal to
(2014 Main)
(a) $\alpha \beta$
(b) $\frac{1}{\alpha \beta}$
(c) 1
(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, $f(n)=\alpha^{n}+\beta^{n}, f(1)=\alpha+\beta, f(2)=\alpha^{2}+\beta^{2}$,
$\begin{aligned} & f(3)=\alpha^{3}+\beta^{3}, f(4)=\alpha^{4}+\beta^{4} \\ \text { Let } & \Delta=\begin{vmatrix}3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4)\end{vmatrix} \\ \Rightarrow \quad \Delta & =\begin{vmatrix}3 & 1+\alpha+\beta & 1+\alpha^{2}+\beta^{2} \\ 1+\alpha+\beta & 1+\alpha^{2}+\beta^{2} & 1+\alpha^{3}+\beta^{3} \\ 1+\alpha^{2}+\beta^{2} & 1+\alpha^{3}+\beta^{3} & 1+\alpha^{4}+\beta^{4}\end{vmatrix} \end{aligned}$
$= \begin{vmatrix}1 \cdot 1+1 \cdot 1+1 \cdot 1 & 1 \cdot 1+1 \cdot \alpha+1 \cdot \beta \\ 1 \cdot 1+1 \cdot \alpha+1 \cdot \beta & 1 \cdot 1+\alpha \cdot \alpha+\beta \cdot \beta \\ 1 \cdot 1+1 \cdot \alpha^{2}+1 \cdot \beta^{2} & 1 \cdot 1+\alpha^{2} \cdot \alpha+\beta^{2} \cdot \beta\end{vmatrix}$ $ \begin{vmatrix} 1 \cdot 1+1 \cdot \alpha^{2}+1 \cdot \beta^{2} \\ 1 \cdot 1+\alpha \cdot \alpha^{2}+\beta \cdot \beta^{2} \\ 1 \cdot 1+\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 $\Delta=(1-\alpha)^{2}(1-\beta)^{2}(\alpha-\beta)^{2}$
$ \text { But given, } \quad \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 \quad K=1 $
