Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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 :

  1. A a<sub>2</sub>a<sub>6</sub> $-$ a<sub>4</sub>a<sub>8</sub>
  2. B a<sub>9</sub>
  3. C a<sub>1</sub>a<sub>9</sub> $-$ a<sub>3</sub>a<sub>7</sub> Correct answer
  4. D a<sub>5</sub>

Solution

${a_r} = {e^{{{i2\pi r} \over 9}}}$, r = 1, 2, 3, ......, a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..... are in G.P.<br><br>$$\left| {\matrix{ {{a_1}} &amp; {{a_2}} &amp; {{a_3}} \cr {{a_n}} &amp; {{a_5}} &amp; {{a_6}} \cr {{a_7}} &amp; {{a_8}} &amp; {{a_9}} \cr } } \right| = \left| {\matrix{ {{a_1}} &amp; {a_1^2} &amp; {a_1^3} \cr {a_1^4} &amp; {a_1^5} &amp; {a_1^6} \cr {a_1^7} &amp; {a_1^8} &amp; {a_1^9} \cr } } \right| $$ <br><br>$$= {a_1}\,.\,a_1^4\,.\,a_1^7\left| {\matrix{ 1 &amp; {{a_1}} &amp; {a_1^2} \cr 1 &amp; {{a_1}} &amp; {a_1^2} \cr 1 &amp; {{a_1}} &amp; {a_1^2} \cr } } \right| = 0$$<br><br>Now, ${a_1}{a_9} - {a_3}{a_7} = {a_1}^{10} - {a_1}^{10} = 0$

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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