A) \[x\]and y
B) \[x\] and z
C) y and z
D) \[x,y\]and z
E) independent of\[x,y\]and z
Correct Answer: E
Solution :
\[\because \]a, b, c, d, e and\[f\]are in GP. \[\therefore \]\[a=a,b=ar,c=a{{r}^{2}},d=a{{r}^{3}},e=a{{r}^{4}}\]and \[f=a{{r}^{5}}.\] \[\therefore \] \[\left| \begin{matrix} {{a}^{2}} & {{d}^{2}} & x \\ {{b}^{2}} & {{e}^{2}} & y \\ {{c}^{2}} & {{f}^{2}} & z \\ \end{matrix} \right|=\left| \begin{matrix} {{a}^{2}} & {{a}^{2}}{{r}^{6}} & x \\ {{a}^{2}}{{r}^{2}} & {{a}^{2}}{{r}^{8}} & y \\ {{a}^{2}}{{r}^{4}} & {{a}^{2}}{{r}^{10}} & z \\ \end{matrix} \right|\] \[={{a}^{4}}{{r}^{6}}\left| \begin{matrix} 1 & 1 & x \\ {{r}^{2}} & {{r}^{2}} & y \\ {{r}^{4}} & {{r}^{4}} & z \\ \end{matrix} \right|=0\] Thus, the given determinant is independent of \[x,\text{ }y\] and z.
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