A) \[3{{\alpha }^{3}}\]
B) \[3({{\alpha }^{3}}+{{\alpha }^{6}}+{{\alpha }^{9}})\]
C) \[3(\alpha +{{\alpha }^{2}}+{{\alpha }^{3}})\]
D) \[3{{\alpha }^{4}}\]
E) \[0\]
Correct Answer: E
Solution :
Let \[\Delta =\left| \begin{matrix} \alpha & {{\alpha }^{3}} & {{\alpha }^{5}} \\ {{\alpha }^{3}} & {{\alpha }^{5}} & \alpha \\ {{\alpha }^{5}} & \alpha & {{\alpha }^{3}} \\ \end{matrix} \right|\] Let\[\alpha =\omega \]\[\therefore \]\[\Delta =\left| \begin{matrix} \omega & {{\omega }^{3}} & {{\omega }^{5}} \\ {{\omega }^{3}} & {{\omega }^{5}} & \omega \\ {{\omega }^{5}} & \omega & {{\omega }^{3}} \\ \end{matrix} \right|=\left| \begin{matrix} \omega & 1 & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & \omega \\ {{\omega }^{2}} & \omega & 1 \\ \end{matrix} \right|\] \[=\omega ({{\omega }^{2}}-{{\omega }^{2}})-1(1-{{\omega }^{3}})+{{\omega }^{2}}(\omega -{{\omega }^{4}})\] \[=\omega (0)-1+{{\omega }^{3}}+{{\omega }^{3}}-{{\omega }^{6}}=-1+2(1)-1\] \[=-2+2=0\]You need to login to perform this action.
You will be redirected in
3 sec