Animated Solution for Mathematics - Matrices and Determinants: If $A=\frac{1}{2}\left[\begin{array}{cc}1& \sqrt{3}\\ -\sqrt{3}& 1\end{array}\right]$, then :

Welcome, future engineer. Today, we are going to dismantle a problem that often intimidates students.

When you see a matrix like

your first instinct might be to reach for a pen and start multiplying $A\times A$. Resist that urge! In the world of JEE Advanced, brute force is rarely the intended path. There is almost always a deeper, more elegant structure waiting to be uncovered.

The Pattern Recognition

Let's distribute that scalar $\frac{1}{2}$ into the matrix. We get

Look at these entries. They are not random. They are the exact values of $\cos (\frac{\pi }{3})$ and $\sin (\frac{\pi }{3})$.

This is the 'Aha!' moment. We are looking at a rotation matrix, defined as

By identifying $\theta =\frac{\pi }{3}$, we have transformed a matrix algebra problem into a problem of circular motion.

The Power of Rotation

Here is the secret weapon: when you raise a rotation matrix to the power $n$, you are essentially applying the rotation $n$ times. Geometrically, this means the angle simply scales by $n$.

Mathematically, this is expressed as

This is a direct consequence of De Moivre's Theorem. Instead of multiplying matrices, we are now just multiplying angles. How much time did we just save?

Conquering the Powers

Let's tackle $A^{30}$. With $n=30$ and $\theta =\frac{\pi }{3}$, the new angle is $30\times \frac{\pi }{3}=10\pi$.

Now, evaluate the matrix:

Since $10\pi$ represents five full revolutions around the unit circle, we are back at the start. Thus, $\cos (10\pi )=1$ and $\sin (10\pi )=0$. We get

The identity matrix!

Next, consider $A^{25}$. Here, $n=25$, so the angle is $25\times \frac{\pi }{3}=\frac{25\pi }{3}$.

We can break this down as $8\pi +\frac{\pi }{3}$. Because sine and cosine have a period of $2\pi$, the $8\pi$ (four full rotations) effectively vanishes.

We are left with $\cos (\frac{\pi }{3})$ and $\sin (\frac{\pi }{3})$, which brings us right back to our original matrix $A$. So, $A^{25}=A$.

The Final Synthesis

We have our components: $A^{30}=I$ and $A^{25}=A$. The problem asks us to evaluate the relationship between these powers.

Looking at the expression $A^{30}+A^{25}-A$, we substitute our findings: $I+A-A$. The $A$ terms cancel out with beautiful precision, leaving us with $I$.

This is the beauty of mathematics. We didn't need to perform a single complex multiplication. We used pattern recognition, geometric intuition, and the properties of periodic functions to slice through the problem.

Keep this mindset—look for the structure, not just the numbers—and you will conquer any problem the JEE throws your way.