A) \[\alpha \beta \]
B) \[\frac{1}{\alpha \beta }\]
C) \[1\]
D) \[-1\]
Correct Answer: C
Solution :
\[\left| \begin{matrix} 1+1+1 & 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} \\ 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} \\ 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} & 1+{{\alpha }^{4}}+{{\beta }^{4}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha }^{2}} & {{\beta }^{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|\times \left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|\]\[={{\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{\alpha }^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|}^{2}}\] \[={{(1-\alpha )}^{2}}{{(\alpha -\beta )}^{2}}{{(\beta -1)}^{2}}\]\[K=1\]
You need to login to perform this action.
You will be redirected in
3 sec
