A) 2
B) -1
C) 1
D) 0
Correct Answer: B
Solution :
Given, \[\left| \begin{matrix} a & {{a}^{2}} & 1+{{a}^{3}} \\ b & {{b}^{2}} & 1+{{b}^{3}} \\ c & {{c}^{2}} & 1+{{c}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+\left| \begin{matrix} a & {{a}^{2}} & {{a}^{3}} \\ b & {{b}^{2}} & {{b}^{3}} \\ c & {{c}^{2}} & {{c}^{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \,\,\,\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|+abc\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow (1+abc)\,\,\,\left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|=0\left( \because \left| \begin{matrix} a & {{a}^{2}} & 1 \\ b & {{b}^{2}} & 1 \\ c & {{c}^{2}} & 1 \\ \end{matrix} \right|\ne 0 \right)\] \[\Rightarrow \] \[1+abc=0\Rightarrow abc=-1\]You need to login to perform this action.
You will be redirected in
3 sec