Let ω = -1/2 + i√3/2. Then the value of the determinant |1 1 1; 1 -1-ω² ω²; 1 ω² ω⁴| is (a) 3ω (b) 3

Q: Why can't I just expand the determinant immediately? It's just a 3x3 matrix.

You absolutely could! But in JEE Advanced, time is your most precious resource. Expanding directly involves complex multiplication and potential algebraic errors. Using properties is the 'pro' way to solve it, and it often leads to a much cleaner, faster result.

Q: I keep getting confused with $\omega^4$. Is there a trick?

Always remember $\omega^3 = 1$. Any power of $\omega$ can be reduced by dividing the exponent by 3 and taking the remainder. So $\omega^4 = \omega^{3+1} = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. It's that simple!

Q: Why did we choose $C_1 \rightarrow C_1 + C_2 + C_3$ specifically?

Look at the rows: $1, 1, 1$ and $1, \omega, \omega^2$ and $1, \omega^2, \omega$. Summing them gives $1+1+1=3$ (for the first column) and $1+\omega+\omega^2=0$ (for the others). It's a classic pattern to create zeros and simplify the determinant!

Question

Let $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. Then the value of the determinant $\begin{vmatrix} 1 & 1 & 1 \ 1 & -1-\omega^2 & \omega^2 \ 1 & \omega^2 & \omega^4 \end{vmatrix}$ is

Options:

(A) (a) $3\omega$

(B) (b) $3\omega(\omega-1)$

(C) (c) $3\omega^2$

(D) (d) $3\omega(1-\omega)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Argand plane, looking at the complex number $\omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$. To the uninitiated, this is just a number. But to a JEE aspirant, this is a gateway to symmetry.

When we plot $1$, $\omega$, and $\omega^2$ on the complex plane, they form a perfect equilateral triangle inscribed in a unit circle. This geometric harmony is the key to unlocking our problem.

### The Golden Rules of $\omega$

Before we even touch the determinant, we must arm ourselves with two fundamental properties. First, $\omega^3 = 1$. This is our 'power-reducer'—it allows us to turn any high power of $\omega$ into something manageable.

Second, $1 + \omega + \omega^2 = 0$. This is our 'zero-maker'—it is the secret weapon for simplifying expressions and determinants. Keep these two tools close; they are the keys to the kingdom.

### The Villain: The Determinant

We are faced with the determinant:

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & -1-\omega^2 & \omega^2 \ 1 & \omega^2 & \omega^4 \end{vmatrix}$$

At first glance, it looks intimidating. But let's break it down. Using our golden rules, we know that $1 + \omega + \omega^2 = 0$, which implies $\omega = -1 - \omega^2$.

We also know $\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega$. Suddenly, the 'villain' is not so scary. The determinant simplifies to:

$$\Delta = \begin{vmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{vmatrix}$$

### The Master Stroke: Column Operations

Now, we could expand this, but why do the hard work when we can be clever? Let's apply the column operation $C_1 ightarrow C_1 + C_2 + C_3$.

Look at what happens to the first column: the first row becomes $1+1+1=3$, the second row becomes $1+\omega+\omega^2=0$, and the third row becomes $1+\omega^2+\omega=0$. Our determinant is now:

$$\Delta = \begin{vmatrix} 3 & 1 & 1 \ 0 & \omega & \omega^2 \ 0 & \omega^2 & \omega \end{vmatrix}$$

### Final Calculation

With two zeros in the first column, the expansion becomes trivial:

$$\Delta = 3 \begin{vmatrix} \omega & \omega^2 \ \omega^2 & \omega \end{vmatrix} = 3(\omega^2 - \omega^4)$$

Since $\omega^4 = \omega$, we get $\Delta = 3(\omega^2 - \omega)$. Factoring out $\omega$, we arrive at the final, elegant answer:

**$\Delta = 3\omega(\omega - 1)$**

This is the beauty of mathematics—taking a complex, messy problem and, through the application of symmetry and logic, reducing it to a simple, clean result. Keep this mindset, and no determinant will ever intimidate you again!

Animated Solution for Mathematics - Complex Numbers: Let $ω = - \frac{1}{2} + i \frac{3}{2}$ . Then the value of the determinant $111 1 - 1 - ω^{2} ω^{2} 1 ω^{2} ω^{4}$ is

Select Answer:

Visualized Solution

Visualizing $ω$ on the Argand Plane

The Cube Roots of Unity

Core Properties of $ω$

The Given Determinant

Simplifying $- 1 - ω^{2}$

Simplifying $ω^{4}$

Rewriting the Determinant

Applying Column Operation

Calculating the New Column

The Transformed Determinant

Expanding Along Column 1

Cross Multiplication

Final Substitution and Factoring

Matching the Option

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Golden Rules of $ω$

The Villain

The Master Stroke

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ω=−21​+i23​​. Then the value of the determinant ​111​1−1−ω2ω2​1ω2ω4​​ is

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Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω on the Argand Plane

The Cube Roots of Unity

Core Properties of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω