Animated Solution for Mathematics - Complex Numbers: If $1,\omega ,{\omega }^2$ are the cube roots of unity, then $\Delta =\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^n& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right|$ is equal to

Unlocking the Determinant

Welcome, future engineers. Today, we are not merely solving a determinant; we are embarking on a journey into the heart of complex numbers and geometric symmetry.

Problems like this are the bread and butter of JEE Advanced. They test not just your ability to calculate, but your ability to see the hidden structure beneath the surface. Let us peel back the layers of this problem together.

The Geometry of Unity

Before we touch the determinant

let us ground ourselves in the complex plane. The cube roots of unity, $1,\omega ,{\omega }^2$, are the solutions to the equation $z^3=1$.

If you were to plot these on the Argand plane, you would see them sitting perfectly on the unit circle, spaced exactly $120^{\circ }$ apart. They form the vertices of an equilateral triangle centered at the origin.

The fundamental property here is that the sum of these roots is zero: $1+\omega +{\omega }^2=0$. This geometric balance is the secret weapon we will use to dismantle the determinant.

The Power of Cyclic Symmetry

Look closely at the matrix. The elements are shifting in a cyclic dance. In the first row, we have $1,{\omega }^n,{\omega }^{2n}$.

In the second row, the elements have shifted, and in the third, they have shifted again. Whenever you see this cyclic behavior in a determinant, your intuition should scream: "Add the columns!"

By applying the operation $C_1\to C_1+C_2+C_3$, we are essentially summing the elements of each row into the first column. The first column becomes $1+{\omega }^n+{\omega }^{2n}$ for every single row.

Suddenly, we have transformed a complex, intimidating matrix into one where the first column is uniform. This is the moment where the chaos turns into order.

The Bifurcation of Cases

Now, we have our determinant in the form:

We have successfully factored out the term $(1+{\omega }^n+{\omega }^{2n})$. The entire value of our determinant now hinges on this multiplier.

But wait—the value of ${\omega }^n$ is not static. It depends entirely on the integer $n$. We must split our analysis into two distinct cases.

If $n$ is not a multiple of $3$, then ${\omega }^n$ is either $\omega$ or ${\omega }^2$. It can never be $1$.

This means the set of terms $\{1,{\omega }^n,{\omega }^{2n}\}$ is simply a permutation of the set $\{1,\omega ,{\omega }^2\}$. Because the sum of the cube roots of unity is zero, the expression $1+{\omega }^n+{\omega }^{2n}$ must also be zero.

If the multiplier is zero, the entire determinant $\Delta$ collapses to zero. It is elegant, it is swift, and it is powerful.

What if $n=3k$? Here, ${\omega }^n=({\omega }^3{)}^k=1^k=1$.

If we substitute this back into our original determinant, every single entry becomes $1$. We get a matrix where every row is identical: $[1,1,1]$.

A fundamental property of determinants is that if any two rows (or columns) are identical, the determinant is zero. Thus, even in this case, the result is zero.

Whether $n$ is a multiple of $3$ or not, the result is the same. The determinant $\Delta$ is zero for all integer values of $n$.

We started with a complex-looking matrix, applied a simple column operation, analyzed the symmetry of the roots of unity, and arrived at a clean, definitive answer. This is the beauty of mathematics—taking a seemingly impossible problem and finding the underlying simplicity.