Question

Suppose $\alpha, \beta$ are two real numbers and $f(n)=\alpha^n+\beta^n$.
Let $\Delta=\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix}$
If $\Delta=k(\alpha-1)^2(\beta-1)^2(\alpha-\beta)^2$, then $k$ is equal to

• A. 1
• B. $4 \alpha \beta$
• C. 9
• D. $\alpha^2 \beta^2$

Answer 1

Answer: (A)

$\Delta =\begin{vmatrix}1+1+1 & 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 & 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}=\Delta_1^2$
where$\Delta_1=\begin{vmatrix}1 & 1 & 1 \\1 & \alpha & \beta \\1 & \alpha^2 & \beta^2\end{vmatrix}$
Applying $C_3 \rightarrow C_3-C_2$ and $C_2 \rightarrow C_2-C_1$, we get
$\Delta_1=\begin{vmatrix}1 & 0 & 0 \\ 1 & \alpha-1 & \beta-\alpha \\ 1 & \alpha^2-1 & \beta^2-\alpha^2\end{vmatrix}$
$=(\alpha-1)(\beta-\alpha)\begin{vmatrix}1 & 1 \\ \alpha+1 & \beta+\alpha\end{vmatrix}$
$=(\alpha-1)(\beta-1)(\beta-\alpha) $
$\text { Thus, } \Delta=(\alpha-1)^2(\beta-1)^2(\alpha-\beta)^2$
$\therefore k=1$