Question

If ω is a cube root of unity, then find the value of the determinant \(\rm \begin{vmatrix} 1 + \omega & \omega^2 & -\omega \\\ 1 + \omega^2 & \omega & -\omega^2 \\\ \omega^2 + \omega & \omega & -\omega^2 \end{vmatrix}\) is
1. 3ω 
2. -3ω
3. 3ω²
4. -3ω²

Answer 1

Correct Answer - Option 4 : -3ω²

Concept:

ω is a cube root of unity.

Property of cube root of unity:

• ω3 = 1
• 1 + ω + ω2  = 0

Calculations:

Given, ω is a cube root of unity.

⇒ω³ = 1, ad 1 + ω + ω²  = 0

Now, consider the determinant

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

Taking ω and -ω common from C₂ and C₃ respectively,

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

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

= \(\rm -ω^2 [-ω^2+1+1-ω]\)

= \(\rm ω -2ω^2+ 1\)

= \(\rm 1+ ω + ω^2-3ω^2\)

= 0 - 3ω2

= - 3ω2

Hence the required value is -3ω2