Question

If $\alpha ,\beta \ne 0$, $f\left(n\right)={\alpha }^n+{\beta }^n$ 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 $1$

Given, $f\left(n\right)={\alpha }^n+{\beta }^n$

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

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

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

$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$

$={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$

$C_2+C_1-C_2,C_3+C_1-C_3$

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

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

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

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

$\therefore k=1$