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 :
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>$$\left| {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{a_n}} & {{a_5}} & {{a_6}} \cr
{{a_7}} & {{a_8}} & {{a_9}} \cr
} } \right| = \left| {\matrix{
{{a_1}} & {a_1^2} & {a_1^3} \cr
{a_1^4} & {a_1^5} & {a_1^6} \cr
{a_1^7} & {a_1^8} & {a_1^9} \cr
} } \right| $$
<br><br>$$= {a_1}\,.\,a_1^4\,.\,a_1^7\left| {\matrix{
1 & {{a_1}} & {a_1^2} \cr
1 & {{a_1}} & {a_1^2} \cr
1 & {{a_1}} & {a_1^2} \cr
} } \right| = 0$$<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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