Animated Solution for Mathematics - Complex Numbers: Let $\omega =-1/2+i\sqrt{3}/2$, then the value of the det. $$\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& {\omega }^4\end{array}\right|$$ is

Imagine you are standing before a complex determinant, a $3\times 3$ grid filled with powers of $\omega$. At first glance, it looks like a chaotic mess of terms. But as a student of mathematics, you know that chaos is often just order in disguise.

Our protagonist is $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, the famous complex cube root of unity. This number is not just a value; it is a key that unlocks a world of symmetry.

Before we even touch the determinant, let us recall the two superpowers of $\omega$. First, ${\omega }^3=1$, which means powers of $\omega$ cycle every three steps. Second, the sum of the roots $1+\omega +{\omega }^2=0$. These two identities are our compass.

Look at the element $a_{22}=-1-{\omega }^2$. Using our second superpower, $1+\omega +{\omega }^2=0$, we can rearrange this to see that $-1-{\omega }^2=\omega$. Suddenly, that complex term becomes a simple $\omega$.

Now, look at the bottom right corner, $a_{33}={\omega }^4$. Using our first superpower, ${\omega }^3=1$, we can write:

The matrix is already starting to look much cleaner. We have transformed a daunting expression into an elegant structure:

This is where the JEE spirit truly shines. We want to create zeros, as they are the best friends of a mathematician when expanding determinants.

Notice what happens if we add all three rows together using the operation $R_1\to R_1+R_2+R_3$:

The first element becomes $1+1+1=3$. The second element becomes $1+\omega +{\omega }^2$, which is $0$. The third element becomes $1+{\omega }^2+\omega$, which is also $0$.

Our determinant now stands as:

Expanding along the first row is now trivial. We have:

Calculating this $2\times 2$ minor, we get $(\omega \cdot \omega )-({\omega }^2\cdot {\omega }^2)={\omega }^2-{\omega }^4$. Since ${\omega }^4=\omega$, this simplifies to $3({\omega }^2-\omega )$.

Factoring out the $\omega$, we arrive at our final, beautiful result:

This journey shows us that even the most intimidating problems can be tamed with the right tools and a bit of patience. Keep practicing, keep exploring, and remember that every complex problem has a simple, elegant solution waiting to be found.