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

The correct option is C $1$

$f\left(1\right)=\alpha +\beta$
$f\left(2\right)={\alpha }^2+{\beta }^2$
$f\left(3\right)={\alpha }^3+{\beta }^3$
$f\left(4\right)={\alpha }^4+{\beta }^4$
So, $\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|$
Splitting it as a product of two determinants.
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
$={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$

$C_1\to C_1-C_2,C_2\to C_2-C_3$
$={\left|\begin{array}{ccc}0& 0& 1\\ 1-\alpha & \alpha -\beta & \beta \\ 1-{\alpha }^2& {\alpha }^2-{\beta }^2& {\beta }^2\end{array}\right|}^2$
$=\left[\right(1-\alpha \left)\right(a-\beta ){]}^2{\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& \beta \\ 1+\alpha & \alpha +\beta & {\beta }^2\end{array}\right|}^2$
$=\left[\right(1-\alpha \left)\right(\alpha -\beta \left){]}^2\right(\beta -1{)}^2$
$=(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
Hence, $K=1$