Animated Solution for Mathematics - Matrices and Determinants: If $\omega (\ne 1)$ is a cube root of unity, then $\left|\begin{array}{ccc}1& 1+i+{\omega }^2& {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|=$

When you first encounter a determinant like this, it is natural to feel a surge of intimidation. You see the imaginary unit $i$, the cube root of unity $\omega$, and a $3\times 3$ grid of complex expressions.

Your instinct might be to start expanding along the first row, but stop! That is exactly what the examiner wants you to do. They want you to get lost in a forest of algebraic terms, hoping you will make a sign error or lose track of a power of $\omega$.

In the world of JEE Advanced, the determinant is not just a calculation; it is a puzzle. The key is to look for the hidden structure before you start the heavy lifting.

Before we touch the determinant, let us remind ourselves of our tools. We have the imaginary unit $i$, where $i^2=-1$.

Then we have $\omega$, the cube root of unity, which satisfies the golden identity:

This identity is your most powerful weapon. Whenever you see these terms, your brain should immediately start looking for ways to group them into this sum. If you can create a $1+\omega +{\omega }^2$ somewhere, you have essentially created a zero.

Why do we choose row operations? Because determinants are invariant under certain transformations. Adding or subtracting rows does not change the value of the determinant, but it can drastically simplify the elements inside.

We are looking for a linear combination of rows that makes the first row vanish. Let us test the operation $R_1\to R_1-R_2+R_3$.

This is a bold move, but it is calculated. We are checking if the rows are linearly dependent.

The Magic of the Operation

Let us perform the operation $R_1\to R_1-R_2+R_3$ element by element.

For the first column, we have $1-(1-i)+(-i)$. Expanding this, we get $1-1+i-i$, which is exactly $0$. The first element is gone!

Now for the second column: $(1+i+{\omega }^2)-(-1)+(-i+\omega -1)$. Simplifying this, the $i$ and $-i$ cancel out, and we are left with:

Since $1+\omega +{\omega }^2=0$, this element also becomes $0$.

Finally, for the third column: ${\omega }^2-({\omega }^2-1)+(-1)$. This becomes ${\omega }^2-{\omega }^2+1-1$, which is also $0$.

We have transformed the entire first row into $[0,0,0]$.

When you see that first row become $[0,0,0]$, you should feel a sense of triumph. The property of determinants states that if any row or column is entirely zero, the determinant itself is zero.

We did not need to expand the determinant; we simply revealed its nature. This is the beauty of mathematics—the most complex problems often have the most elegant solutions.

Keep this mindset, and you will conquer any determinant the JEE throws at you. The final answer is 0.