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  4. | beginmatrix 1 2 3 13 23 33 15 25 35 endmatrix | equal to

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Q. $ \left| \begin{matrix} 1 & 2 & 3 \\ {{1}^{3}} & {{2}^{3}} & {{3}^{3}} \\ {{1}^{5}} & {{2}^{5}} & {{3}^{5}} \\ \end{matrix} \right| $ equal to

J & K CETJ & K CET 2006

Solution:

Let $ A=\left| \begin{matrix} 1 & 2 & 3 \\ {{1}^{3}} & {{2}^{3}} & {{3}^{3}} \\ {{1}^{5}} & {{2}^{5}} & {{3}^{5}} \\ \end{matrix} \right| $
$ =1.2.3\left| \begin{matrix} 1 & 1 & 1 \\ {{1}^{2}} & {{2}^{2}} & {{3}^{2}} \\ {{1}^{4}} & {{2}^{4}} & {{3}^{4}} \\ \end{matrix} \right| $
$ =6\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 4 & 9 \\ 1 & 16 & 81 \\ \end{matrix} \right| $
$ =6\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 3 & 5 \\ 1 & 15 & 65 \\ \end{matrix} \right|({{C}_{2}}\to {{C}_{2}}-{{C}_{1}},{{C}_{3}}\to {{C}_{3}}-{{C}_{2}}) $
$ =6.3.5\left| \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 5 & 13 \\ \end{matrix} \right| $
$ =90[1(13-5)] $
$ =90\times 8 $
$ =720=6! $