If A = [[cos θ, -sin θ], [sin θ, cos θ]], then the matrix A⁻⁵⁰ when θ = π/12, is equal to : (A) [[√3

Q: Wait, can't I just multiply the matrix 50 times? It's just arithmetic, right?

Technically, yes, you could attempt to multiply the matrix 50 times, but in the context of a JEE Advanced exam, time is your most precious currency. The beauty of this problem lies in recognizing the pattern. If you try to brute-force it, you will likely make a sign error or run out of time. Always look for the underlying geometric property first!

Q: Why does the negative power flip the sign of the sine terms?

Think of the negative power as the inverse operation. If $A$ rotates a vector counter-clockwise by $\theta$, then $A^{-1}$ rotates it clockwise by $\theta$. When you raise it to the power of $-50$, you are effectively rotating the vector clockwise by $50\theta$. Mathematically, this manifests as $\sin(-50\theta)$, which, because sine is an odd function, becomes $-\sin(50\theta)$. It is consistent and elegant!

Q: How do I know when to stop simplifying the angle $50\theta$?

Great question! You stop when the angle is within the principal range $[0, 2\pi)$ or when it is a standard angle like $\frac{\pi}{6}, \frac{\pi}{4}, \text{ or } \frac{\pi}{3}$. In this case, we reduced $\frac{50\pi}{12}$ to $4\pi + \frac{\pi}{6}$. Since $4\pi$ represents two full rotations, it brings us back to the same position as $\frac{\pi}{6}$. Always look for these full-circle multiples to simplify your life.

Question

If $A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$, then the matrix $A^{-50}$ when $	heta = \frac{\pi}{12}$, is equal to :

Options:

(A) $\begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$

(B) $\begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}$

(C) $\begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}$

(D) $\begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix}$

Sigma Percentile · Accepted Answer

### The Geometry of Rotation: Unlocking the Matrix A

Welcome, future engineer. Today, we are not just solving a matrix problem; we are embarking on a journey into the heart of linear algebra.

When you first look at the matrix

$$A = \begin{bmatrix} \cos 	heta & -\sin 	heta \ \sin 	heta & \cos 	heta \end{bmatrix}$$

I want you to stop seeing it as a grid of numbers. I want you to see it as a machine.

In the world of physics and computer graphics, this is the 'Rotation Matrix'. It is the engine that turns objects in space. If you take a vector $\vec{v}$ and multiply it by $A$, you are physically rotating that vector by an angle $	heta$ in the counter-clockwise direction.

This geometric intuition is your greatest weapon.

### The Power of Powers: Avoiding the Trap

Now, the problem asks us to find $A^{-50}$. A student who hasn't mastered the theory might panic, thinking they must multiply this matrix by itself 50 times.

If you try that, you will be lost in a sea of trigonometric identities and arithmetic errors. But you are smarter than that.

You know the property of rotation matrices: if applying $A$ once rotates a vector by $	heta$, then applying it $n$ times is equivalent to rotating it by $n	heta$. Therefore:

$$A^n = \begin{bmatrix} \cos(n	heta) & -\sin(n	heta) \ \sin(n	heta) & \cos(n	heta) \end{bmatrix}$$

This is the 'Aha!' moment. We have just reduced a terrifying matrix exponentiation problem into a simple multiplication of an angle.

### The Negative Power: A Clockwise Shift

We are dealing with $A^{-50}$. The negative sign might look intimidating, but it is just a direction indicator.

If a positive power is a counter-clockwise rotation, a negative power is simply a clockwise rotation. When we substitute $n = -50$ into our general formula, we get:

$$A^{-50} = \begin{bmatrix} \cos(-50	heta) & -\sin(-50	heta) \ \sin(-50	heta) & \cos(-50	heta) \end{bmatrix}$$

Now, we must be precise. We invoke the even and odd properties of trigonometry.

We know that $\cos(-x) = \cos(x)$ because cosine is an even function. Conversely, $\sin(-x) = -\sin(x)$ because sine is an odd function.

Applying this, our matrix becomes:

$$A^{-50} = \begin{bmatrix} \cos(50	heta) & \sin(50	heta) \ -\sin(50	heta) & \cos(50	heta) \end{bmatrix}$$

### The Final Reduction: Bringing it Home

We are given $	heta = \frac{\pi}{12}$. Our angle is $50	heta = 50 	imes \frac{\pi}{12} = \frac{50\pi}{12}$.

Let's simplify this fraction. Dividing both numerator and denominator by 2, we get $\frac{25\pi}{6}$.

Now, we use the periodicity of trigonometric functions. We can write this as:

$$\frac{25\pi}{6} = \frac{24\pi + \pi}{6} = 4\pi + \frac{\pi}{6}$$

That $4\pi$ is just two full revolutions. It brings us right back to where we started.

So, $\cos(4\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6})$ and $\sin(4\pi + \frac{\pi}{6}) = \sin(\frac{\pi}{6})$.

We know these values by heart: $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.

Substituting these back into our matrix, we arrive at the final result:

$$A^{-50} = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$$

Look at that elegance. We didn't fight the matrix; we understood its nature, and it yielded the answer to us. This is the power of conceptual clarity in JEE Advanced.

Animated Solution for Mathematics - Matrices and Determinants: If $A = [cos θ sin θ - sin θ cos θ]$ , then the matrix $A^{- 50}$ when $θ = \frac{π}{12}$ , is equal to :

Select Answer:

Visualized Solution

Identify the Matrix $A$

Power Property of Rotation Matrices

Applying the Property for $A^{- 50}$

Simplifying Negative Angles

Substituting $θ = \frac{π}{12}$

Simplifying the Angle

Evaluating Trigonometric Values

Final Matrix Assembly

The Sigma Insight: Algebraic Operations on Matrices

The Geometry of Rotation

The Power of Powers

The Negative Power

The Final Reduction

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A=[cosθsinθ​−sinθcosθ​], then the matrix A−50 when θ=12π​, is equal to :