Let ω = -1/2 + idfrac√(3)2, then value of the determinant beginvmatrix 1 & 1 & 1 \ 1 & -1-ω^2 & ω^2 \ 1 & ω^2 & ω endvmatrix is

Let ω = -1/2 + idfrac√(3)2, then value of the determinant… — JEE Main 2002 Mathematics PYQ

AIEEE 2002

Let

\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}

, then value of the determinant

\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}

A.
$3\omega$
B.
$3\omega(\omega - 1)$
C.
$3\omega^2$
D.
$3\omega(1 - \omega)$

Show answer and solution

Correct option: B

Solution

ω is a primitive cube root of unity, so 1 + ω + ω² = 0, meaning -1 - ω² = ω.

The determinant becomes:
|1 1 1 |
|1 ω ω² |
|1 ω² ω |

Apply C2 → C2 - C1 and C3 → C3 - C1:
|1 0 0 |
|1 ω-1 ω²-1 |
|1 ω²-1 ω-1 |

Expanding along row 1: (ω-1)(ω-1) - (ω²-1)(ω²-1) = (ω-1)² - (ω²-1)²

Factor as difference of squares: [(ω-1) - (ω²-1)][(ω-1) + (ω²-1)] = (ω - ω²)(ω + ω² - 2)

Since ω + ω² = -1: ω + ω² - 2 = -3

So the value = (ω - ω²)(-3) = 3(ω² - ω)

Now factor: 3(ω² - ω) = 3ω(ω - 1)·(−1)... let me recheck: 3(ω² - ω) = 3ω(ω/ω · ω - 1) = 3ω(ω - 1).

Indeed ω² - ω = ω(ω - 1), so the determinant = **3

[AIEEE 2002]

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