Animated Solution for Mathematics - Matrices and Determinants: If $A=\left(\begin{array}{cc}\cos \theta & i\sin \theta \\ i\sin \theta & \cos \theta \end{array}\right)$, $(\theta =\frac{\pi }{24})$ and $A^5=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ where $i=\sqrt{-1}$, then which one of the following is not true?

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a matrix problem; we are uncovering a hidden symmetry.

When you first look at the matrix

it might look like a standard, albeit slightly intimidating, grid of numbers. But I want you to see it as a transformation.

When we are asked to find $A^5$, the instinct might be to dive into brute-force multiplication. But stop. Take a breath.

In the world of competitive physics and mathematics, brute force is rarely the intended path. Let us look for the rhythm in the numbers.

Unveiling the Pattern

Instead of jumping to the fifth power, let us test the second power. We calculate $A^2=A\cdot A$.

When we multiply the first row by the first column, we get $(\cos \theta )(\cos \theta )+(i\sin \theta )(i\sin \theta )$. This simplifies to ${\cos }^2\theta +i^2{\sin }^2\theta$.

Since $i^2=-1$, this becomes ${\cos }^2\theta -{\sin }^2\theta$. Does that ring a bell? It is the classic double-angle identity for $\cos 2\theta$!

Now, look at the top-right element: $(\cos \theta )(i\sin \theta )+(i\sin \theta )(\cos \theta )=2i\sin \theta \cos \theta$. Again, we see the double-angle identity, this time for $\sin 2\theta$.

So, $A^2$ becomes

The pattern is clear: the power of the matrix simply acts as a multiplier for the angle $\theta$. By induction, we can confidently state that

With our general form $A^n$ in hand, finding $A^5$ is trivial. We simply set $n=5$ to get

Comparing this to the given matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, we identify $a=\cos 5\theta$, $b=i\sin 5\theta$, $c=i\sin 5\theta$, and $d=\cos 5\theta$.

Now, we test the options. Option (A) claims $a^2-b^2=\frac{1}{2}$.

Let us check:

Since $1 eq \frac{1}{2}$, we have found our false statement!

I know that sometimes these problems feel like a series of hoops to jump through. But look at what we just did.

We took a complex-looking matrix, found its underlying geometric soul, and used that to dismantle the options with surgical precision.

This is the essence of the JEE Advanced mindset: observation, generalization, and verification. Keep practicing this, and you will find that even the most daunting problems start to reveal their secrets to you.