A) \[abc\]
B) 0
C) 1
D) \[a+b+c\]
Correct Answer: B
Solution :
\[\left| \begin{matrix} 1/a & {{a}^{2}} & bc \\ 1/b & {{b}^{2}} & ca \\ 1/c & {{c}^{2}} & ab \\ \end{matrix} \right|=\frac{1}{abc}\left| \begin{matrix} 1 & {{a}^{3}} & abc \\ 1 & {{b}^{3}} & abc \\ 1 & {{c}^{3}} & abc \\ \end{matrix} \right|\] \[=\frac{abc}{abc}\left| \begin{matrix} 1 & {{a}^{3}} & 1 \\ 1 & {{b}^{3}} & 1 \\ 1 & {{c}^{3}} & 1 \\ \end{matrix} \right|\] \[=0\] [\[\because \]\[{{C}_{1}}\]and\[{{C}_{3}}\]are same]
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