A) \[8\]
B) \[16\]
C) \[32\]
D) \[64\]
Correct Answer: D
Solution :
[d] Let \[D=\left| \begin{matrix} {{(\beta +\gamma -\alpha -\delta )}^{4}} & {{(\beta +\gamma -\alpha -\delta )}^{2}} & 1 \\ {{(\gamma +\alpha -\beta -\delta )}^{4}} & {{(\gamma +\alpha -\beta -\delta )}^{2}} & 1 \\ {{(\alpha +\beta -\gamma -\delta )}^{4}} & {{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 \\ \end{matrix} \right|\] Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}},\] we get \[D=\left| \begin{matrix} {{(\beta +\gamma -\alpha -\delta )}^{4}}-{{(\alpha +\beta -\gamma -\delta )}^{4}} & {{(\beta +\gamma -\alpha -\delta )}^{2}}-{{(\alpha +\beta -\gamma -\delta )}^{2}} & 0 \\ {{(\gamma +\alpha -\beta -\delta )}^{4}}-{{(\alpha +\beta -\gamma -\delta )}^{4}} & {{(\gamma +\alpha -\beta -\delta )}^{2}}-{{(\alpha +\beta -\gamma -\delta )}^{2}} & 0 \\ {{(\alpha +\beta -\gamma -\delta )}^{4}} & {{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 \\ \end{matrix} \right|\]\[4(\beta -\delta )(\gamma -\alpha )\times 4(\alpha -\delta )(\gamma -\beta )\times \] \[\left| \begin{matrix} {{(\beta +\gamma -\alpha -\delta )}^{2}}+{{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 & 0 \\ {{(\gamma +\alpha -\beta -\delta )}^{2}}+{{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 & 0 \\ {{(\alpha +\beta -\gamma -\delta )}^{2}} & {{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 \\ \end{matrix} \right|\]Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}},\] we get \[D=16(\beta -\delta )(\gamma -\alpha )(\alpha -\delta )(\gamma -\beta ).4(\gamma -\delta )(\beta -\alpha )\times \]\[\left| \begin{matrix} 1 & 0 & 0 \\ {{(\gamma +\alpha -\beta -\delta )}^{2}}+{{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 & 0 \\ {{(\alpha +\beta -\gamma -\delta )}^{4}} & {{(\alpha +\beta -\gamma -\delta )}^{2}} & 1 \\ \end{matrix} \right|\]\[=-64(\alpha -\beta )(\alpha -\gamma )(\alpha -\delta )(\beta -\gamma )(\beta -\delta )(\gamma -\delta )\]
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