• # question_answer If$\alpha ,\beta \ne 0,$and$f(n)={{\alpha }^{n}}+{{\beta }^{n}}$and$\left| \begin{matrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \\ \end{matrix} \right|=K$${{(1-\alpha )}^{2}}{{(1-\beta )}^{2}}{{(\alpha -\beta )}^{2}},$then K is equal to:   JEE Main  Solved  Paper-2014 A) $\alpha \beta$                                              B) $\frac{1}{\alpha \beta }$ C) $1$             D) $-1$

$\left| \begin{matrix} 1+1+1 & 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} \\ 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} \\ 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} & 1+{{\alpha }^{4}}+{{\beta }^{4}} \\ \end{matrix} \right|$ $=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha }^{2}} & {{\beta }^{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|$ $=\left| \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|$$={{\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|}^{2}}$ $={{(1-\alpha )}^{2}}{{(\alpha -\beta )}^{2}}{{(\beta -1)}^{2}}$$K=1$