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Matrices

Question
CBSEENMA12036130
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Let  ω be a complex number such that 2ω +1 = z where z = √-3. if

open vertical bar table row 1 1 1 row 1 cell negative straight omega squared end cell cell straight omega squared end cell row 1 cell straight omega squared end cell cell straight omega to the power of 7 end cell end table close vertical bar space equals space 3 straight k
then k is equal to 

  • 1

  • -z

  • z

  • -1

Solution

B.

-z

Here ω is complex cube of unity
straight R subscript 1 rightwards arrow with space on top straight R subscript 1 space plus straight R subscript 2 space plus space straight R subscript 3 space equals space open vertical bar table row 1 0 0 row 1 cell negative straight omega squared minus 1 end cell cell straight omega squared end cell row 1 cell straight omega squared end cell straight omega end table close vertical bar space equals space 3 left parenthesis negative 1 minus straight omega minus straight omega right parenthesis space equals space minus space 3 straight z rightwards double arrow space straight k space equals space minus straight z

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