Question

If $f\left(n\right)={\alpha }^n+{\beta }^n$ 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. 2
• D. -2

Answer 1

The correct option is A 1

$\Delta =\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
Applying $C_2\to -C_2-C_3\to C_3-C_1$,
${\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -1\\ 1& {\alpha }^2-1& {\beta }^2-1\end{array}\right|}^2=(\alpha -1{)}^2(\beta -1{)}^2(\beta -\alpha {)}^2=(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
Hence, k=1