यदि ω एकत्व का एक घन मूल है तो सारणिक \(\rm \begin{vmatrix} 1 + \omega & \omega^2 & -\omega \\\ 1 + \omega^2 & \omega & -\omega^2 \\\ \omega^2 + \omega & \omega & -\omega^2 \end{vmatrix}\) का मान क्या है?

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  1. 3ω 
  2. -3ω
  3. 2
  4. -3ω2

Answer (Detailed Solution Below)

Option 4 : -3ω2
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NIMCET 2020 Official Paper
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संकल्पना:

ω एकत्व का घन मूल है।

एकत्व के घनमूल का गुण:

  • ω = 1
  • 1 + ω + ω= 0

गणना:

दिया, ω एकत्व का घन मूल है।

⇒ω= 1, ad 1 + ω + ω2 = 0

अब, सारणिक

 \(\rm \begin{vmatrix} 1 + ω & ω^2 & -ω \\\ 1 + ω^2 & ω & -ω^2 \\\ ω^2 + ω & ω & -ω^2 \end{vmatrix}\)

ω और -ω को क्रमशः C2 और C3 से उभयनिष्ठ लेकर

= \(\rm-ω^2 \begin{vmatrix} 1 + ω & ω & 1\\\ 1 + ω^2 &1 & ω \\\ ω^2 + ω & 1& ω \end{vmatrix}\)

= \(\rm -ω^2 [(1+ω)(ω-ω)-ω(ω +1-1-ω^2)+(1+ω^2-ω^2-ω)]\)

= \(\rm -ω^2 [-ω^2+1+1-ω]\)

= \(\rm ω -2ω^2+ 1\)

= \(\rm 1+ ω + ω^2-3ω^2\)

= 0 - 3ω 2

= -3ω2

इसलिए, सारणिक का मान -3ω2 है

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