Question

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

• A. $1$
• B. $-1$
• C. $\alpha \beta$
• D. $\alpha \beta \gamma$

Answer 1

The correct option is A $1$

$LHS=\left|\begin{array}{ccc}3& 1+f\left(1\right)& 1+f\left(2\right)\\ 1+f\left(1\right)& 1+f\left(2\right)& 1+f\left(3\right)\\ 1+f\left(2\right)& 1+f\left(3\right)& 1+f\left(4\right)\end{array}\right|$
$=\left|\begin{array}{ccc}& 3& \\ & 1+\alpha +\beta & \\ & 1+{\alpha }^2+{\beta }^2& \end{array}\begin{array}{ccc}& 1+\alpha +\beta & \\ & 1+{\alpha }^2+{\beta }^2& \\ & 1+{\alpha }^3+{\beta }^3& \end{array}\begin{array}{ccc}& 1+{\alpha }^2+{\beta }^2& \\ & 1+{\alpha }^3+{\beta }^3& \\ & 1+{\alpha }^4+{\beta }^4& \end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\times \left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|$
${=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2$
$C_2\to C_2-C_{1}and{C}_3\to C_3-C_1$
${\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -1\\ 1& {\alpha }^2-1& {\beta }^2-1\end{array}\right|}^2$
$=(1-\alpha {)}^2(1-\beta {)}^2{\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& \alpha +1& \beta +1\end{array}\right|}^2$
$=(1-\alpha {)}^2(1-\beta {)}^2(\alpha -\beta {)}^2$
Comparing with the RHS of given expression
Hence $k=1$