Question

If $\alpha ,\beta \ne 0$, and $f\left(n\right)={\alpha }^n+{\beta }^n$ and $\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|=K(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$, then $K$ is equal to

• A. $\alpha \beta$
• B. $\frac{1}{\alpha \beta }$
• C. $1$
• D. $-1$

Answer 1

Answer: (C)

$1$

$\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & {\alpha }^2\\ 1& \beta & {\beta }^2\end{array}\right|$
$={\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -1\\ 1& {\alpha }^2-1& {\beta }^2-1\end{array}\right|}^2$
$=\left(\right(\alpha -1\left)\right({\beta }^2-1)-(\beta -1\left)\right({\alpha }^2-1){)}^2$

$=(\alpha -1{)}^2(\beta -1{)}^2(\alpha -\beta {)}^2\Rightarrow k=1$