JEE MAIN - Mathematics (2021 - 31st August Morning Shift - No. 15)
If $${a_r} = \cos {{2r\pi } \over 9} + i\sin {{2r\pi } \over 9}$$, r = 1, 2, 3, ....., i = $$\sqrt { - 1} $$, then
the determinant $$\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right|$$ is equal to :
the determinant $$\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_4}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right|$$ is equal to :
a2a6 $$-$$ a4a8
a9
a1a9 $$-$$ a3a7
a5
Explanation
$${a_r} = {e^{{{i2\pi r} \over 9}}}$$, r = 1, 2, 3, ......, a1, a2, a3, ..... are in G.P.
$$\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_n}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right| = \left| {\matrix{ {{a_1}} & {a_1^2} & {a_1^3} \cr {a_1^4} & {a_1^5} & {a_1^6} \cr {a_1^7} & {a_1^8} & {a_1^9} \cr } } \right| $$
$$= {a_1}\,.\,a_1^4\,.\,a_1^7\left| {\matrix{ 1 & {{a_1}} & {a_1^2} \cr 1 & {{a_1}} & {a_1^2} \cr 1 & {{a_1}} & {a_1^2} \cr } } \right| = 0$$
Now, $${a_1}{a_9} - {a_3}{a_7} = {a_1}^{10} - {a_1}^{10} = 0$$
$$\left| {\matrix{ {{a_1}} & {{a_2}} & {{a_3}} \cr {{a_n}} & {{a_5}} & {{a_6}} \cr {{a_7}} & {{a_8}} & {{a_9}} \cr } } \right| = \left| {\matrix{ {{a_1}} & {a_1^2} & {a_1^3} \cr {a_1^4} & {a_1^5} & {a_1^6} \cr {a_1^7} & {a_1^8} & {a_1^9} \cr } } \right| $$
$$= {a_1}\,.\,a_1^4\,.\,a_1^7\left| {\matrix{ 1 & {{a_1}} & {a_1^2} \cr 1 & {{a_1}} & {a_1^2} \cr 1 & {{a_1}} & {a_1^2} \cr } } \right| = 0$$
Now, $${a_1}{a_9} - {a_3}{a_7} = {a_1}^{10} - {a_1}^{10} = 0$$
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