A) 2
B) \[-2\]
C) \[-1\]
D) 1
Correct Answer: C
Solution :
\[\left| \begin{matrix} {{x}^{n}} & {{x}^{n+2}} & {{x}^{n+3}} \\ {{y}^{n}} & {{y}^{n+2}} & {{y}^{n+3}} \\ {{z}^{n}} & {{z}^{n+2}} & {{z}^{n+3}} \\ \end{matrix} \right|\] \[=(y-z)(z-x)(x-y)\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\] The degree of determinant \[=n+(n+2)+(n+3)\] \[=3n+5\] and the degree of \[\text{RHS = 2}\] \[\therefore \] \[3n+5=2\] \[\Rightarrow \] \[n=-1.\]You need to login to perform this action.
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