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

Answer 1

The correct option is A $1$

$\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|$

$\alpha +\beta =f\left(1\right){\alpha }^2+{\beta }^2=f\left(2\right){\alpha }^3+{\beta }^3=f\left(3\right){\alpha }^n+{\beta }^n=f\left(n\right)$

$△=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$

Applying $C_2⟶C_2-C_1$ and $C_3⟶C_3-C_1$

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