Let x, y, z 0 are respectively 2^{nd}, 3 | vedclass

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Question

$\Delta  = {{m{x}}^k}{{m{y}}^k}{{m{z}}^{m{k}}}\left| {\begin{array}{*{20}{c}}
1&{{m{ar}}}&{{{m{a}}^2}{{m{r}}^2}}\
1&

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

$\Delta  = {{m{x}}^k}{{m{y}}^k}{{m{z}}^{m{k}}}\left| {\begin{array}{*{20}{c}}
1&{{m{ar}}}&{{{m{a}}^2}{{m{r}}^2}}\
1&{{m{a}}{{m{r}}^2}}&{{{m{a}}^2}{{m{r}}^4}}\
1&{{m{a}}{{m{r}}^3}}&{{{m{a}}^2}{{m{r}}^4}}
\end{array}} ight|$

$ = {{m{a}}^{3{m{k}}}} \cdot {{m{r}}^{6k}} \cdot {{m{a}}^3}{{m{r}}^3}\left| {\begin{array}{*{20}{c}}
1&1&1\
1&{m{r}}&{{{m{r}}^2}}\
1&{{{m{r}}^2}}&{{{m{r}}^4}}
\end{array}} ight|$

$=a^{3(k+1)} \cdot r^{6 k+3}(1-r)\left(r-r^{2}ight)\left(r^{2}-1ight)$

clearly, $\mathrm{k}=-1$

$	herefore \Delta=r^{-2}(1-r)^{2}\left(r^{2}-1ight)=(r-1)^{2}\left(1-\frac{1}{r^{2}}ight)$

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