A) \[a\beta \]
B) \[\frac{1}{a\beta }\]
C) 1
D) \[-\,1\]
Correct Answer: C
Solution :
| \[\left| \begin{matrix} 3 & 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} \\ 1+\alpha +\beta & 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} \\ 1+{{\alpha }^{2}}+{{\beta }^{2}} & 1+{{\alpha }^{3}}+{{\beta }^{3}} & 1+{{\alpha }^{4}}+{{\beta }^{4}} \\ \end{matrix} \right|\] |
| \[=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & {{\alpha }^{2}} & {{\beta }^{2}} \\ \end{matrix} \right|\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \alpha & {{a}^{2}} \\ 1 & \beta & {{\beta }^{2}} \\ \end{matrix} \right|={{\left| \begin{matrix} 1 & 0 & 0 \\ 1 & \alpha -1 & \beta -1 \\ 1 & {{\alpha }^{2}}-1 & {{\beta }^{2}}-1 \\ \end{matrix} \right|}^{2}}\] |
| \[=(\alpha -1)({{\beta }^{2}}-1)-(\beta -1){{({{\alpha }^{2}}-1)}^{2}}\] |
| \[={{(\alpha -1)}^{2}}{{(\beta -1)}^{2}}{{(\alpha -\beta )}^{2}}\] \[\Rightarrow \] \[k=1.\] |
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