A) \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
B) \[{{a}^{2}}b-{{b}^{2}}c\]
C) 0
D) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]
Correct Answer: C
Solution :
We have \[\left| \,\begin{matrix} a & b{{\omega }^{2}} & a\omega \\ b\omega & c & b{{\omega }^{2}} \\ c{{\omega }^{2}} & a\omega & c \\ \end{matrix}\, \right|\] = \[\left| \,\begin{matrix} a(1+\omega ) & b{{\omega }^{2}} & a\omega \\ b(\omega +{{\omega }^{2}}) & c & b{{\omega }^{2}} \\ c({{\omega }^{2}}+1) & a\omega & c \\ \end{matrix}\, \right|\] , \[\{{{C}_{1}}\to {{C}_{1}}+{{C}_{3}}\}\] = \[\left| \,\begin{matrix} -a{{\omega }^{2}} & b{{\omega }^{2}} & a\omega \\ -b & c & b{{\omega }^{2}} \\ -c\omega & a\omega & c \\ \end{matrix}\, \right|\]\[={{\omega }^{2}}\omega \,\,\left| \,\begin{matrix} -a & b & a{{\omega }^{2}} \\ -b & c & b{{\omega }^{2}} \\ -c & a & c{{\omega }^{2}} \\ \end{matrix}\, \right|\] = \[{{\omega }^{2}}\left| \,\begin{matrix} -a & b & a \\ -b & c & b \\ -c & a & c \\ \end{matrix}\, \right|\]= \[-{{\omega }^{2}}\,\left| \,\begin{matrix} a & b & a \\ b & c & b \\ c & a & c \\ \end{matrix}\, \right|\,=0\].
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