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Maths-

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Easy

Question

Let $omega$ = – $fraction numerator 1 over denominator 2 end fraction plus i fraction numerator square root of 3 over denominator 2 end fraction$ . Then the value of the determinant $open vertical bar table row 1 1 1 row 1 cell negative 1 minus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row 1 cell omega to the power of 2 end exponent end cell cell omega to the power of 4 end exponent end cell end table close vertical bar$ is

3 $ω$
3 $ω$ ( $ω$ –1)
3 $ω^{2}$
3 $ω$ (1 – $ω$ )

hint Hint:

We know that, $omega$ is the cube root of unity.
Hence, $1 plus omega plus omega squared equals 0$ and $omega to the power of 3 n end exponent equals 1$ ; $omega to the power of 3 n plus 1 end exponent equals omega$ and $omega to the power of 3 n plus 2 end exponent equals omega$ .

The correct answer is: 3 $omega$ ( $omega$ –1)

Step by step solution:
$open vertical bar table row 1 1 1 row 1 cell negative 1 minus omega to the power of 2 end exponent end cell cell omega to the power of 2 end exponent end cell row 1 cell omega to the power of 2 end exponent end cell cell omega to the power of 4 end exponent end cell end table close vertical bar$
First we open the determinant,
$equals 1 left square bracket open parentheses negative 1 minus omega squared close parentheses omega to the power of 4$ $negative open parentheses omega squared close parentheses open parentheses omega squared close parentheses right square bracket$ $negative 1 left square bracket omega to the power of 4 minus omega squared right square bracket$ $plus 1 left square bracket omega squared minus 1 open parentheses negative 1 minus omega squared close parentheses right square bracket$
$equals negative omega to the power of 4 minus omega to the power of 6 minus omega to the power of 4 minus$ $omega to the power of 4 plus omega squared plus omega squared plus 1 plus omega squared$
$equals negative omega minus 1 minus omega minus omega$ $plus omega squared plus omega squared plus 1 plus omega squared$
$equals negative 3 omega plus 3 omega squared$
$equals 3 omega left parenthesis omega minus 1 right parenthesis$
Hence, option(d) is the correct option.

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