Let A = [[cos α, -sin α], [sin α, cos α]], (α ∈ R) such that A³² = [[0, -1], [1, 0]]. Then a value o

Q: Wait, why do we treat this as a rotation matrix? Can't we just multiply the matrix by itself 32 times?

You could, but that would be a path to madness! In JEE Advanced, we look for patterns and properties. Recognizing the rotation matrix structure is a 'pro-tip' that turns a 20-minute calculation into a 30-second conceptual step. Always look for the underlying geometry before diving into brute-force algebra.

Q: Why is the target matrix $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ equivalent to a $90^\circ$ rotation?

Think of the unit vector $\hat{i} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. If you multiply this vector by the target matrix, you get $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$, which is $\hat{j}$. Since $\hat{i}$ has moved to $\hat{j}$, it has clearly rotated by $90^\circ$ or $\frac{\pi}{2}$ radians. It is all about visualizing the transformation!

Q: Is $32\alpha = \frac{\pi}{2}$ the only possible solution?

For the principal value, yes. In a more general sense, the angle could be $2n\pi + \frac{\pi}{2}$, but since we are looking for a specific value of $\alpha$ that matches our options, the principal value is exactly what we need. Always check your options to see if they align with the principal branch.

Question

Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \ \sin \alpha & \cos \alpha \end{bmatrix}, (\alpha \in R)$ such that $A^{32} = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$. Then a value of $\alpha$ is

Options:

(A) $\frac{\pi}{16}$

(B) $0$

(C) $\frac{\pi}{32}$

(D) $\frac{\pi}{64}$

Sigma Percentile · Accepted Answer

### The Geometry of Matrices: A Journey into Rotation

My dear student, welcome to a moment of mathematical clarity. Often, when we see a matrix like

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

our instinct is to panic or reach for the calculator. But I want you to pause and look at the structure.

Do you see the symmetry? The cosines on the diagonal and the sines on the off-diagonal with that elegant negative sign represent a geometric machine. This is the standard two-dimensional rotation matrix, the mathematical embodiment of a pivot.

### The Power of Iteration

Imagine you are standing on a coordinate plane, holding a vector $V_0$ that rests peacefully on the positive x-axis. When you multiply this vector by matrix $A$, you are essentially giving it a gentle push, rotating it counter-clockwise by an angle $\alpha$.

If you perform this operation $n$ times, you are simply rotating the vector by a total angle of $n\alpha$. This is the beautiful, intuitive power of the rotation matrix:

$$A^n = \begin{bmatrix} \cos n\alpha & -\sin n\alpha \ \sin n\alpha & \cos n\alpha \end{bmatrix}$$

We have bypassed the tedious matrix multiplication and arrived at the heart of the transformation.

### Bridging Algebra and Geometry

Now, let us apply this to our problem. We are given

$$A^{32} = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$$

Using our newfound superpower, we know that $A^{32}$ must be equivalent to a rotation matrix with an angle of $32\alpha$.

So, we write:

$$A^{32} = \begin{bmatrix} \cos 32\alpha & -\sin 32\alpha \ \sin 32\alpha & \cos 32\alpha \end{bmatrix}$$

Let us test the target matrix $\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ on our unit vector $\begin{bmatrix} 1 \ 0 \end{bmatrix}$. The result is $\begin{bmatrix} 0 \ 1 \end{bmatrix}$, which is the y-axis. We have moved from the x-axis to the y-axis, which is a perfect $90^\circ$ or $\frac{\pi}{2}$ radian rotation.

### The Final Resolution

We are now at the finish line. We have equated our theoretical rotation, $32\alpha$, with the geometric reality of $\frac{\pi}{2}$. This gives us the simple, elegant equation:

$$32\alpha = \frac{\pi}{2}$$

To find $\alpha$, we simply divide both sides by $32$. Thus,

$$\alpha = \frac{\pi}{2 	imes 32}$$

which simplifies beautifully to **$\alpha = \frac{\pi}{64}$**.

We did not need to perform complex row reductions or determinant expansions. We simply understood the soul of the matrix. This is the essence of JEE Advanced mathematics—seeing the geometry hidden within the algebra.

Animated Solution for Mathematics - Matrices and Determinants: Let $A = [cos α sin α - sin α cos α], (α \in R)$ such that $A^{32} = [01 - 1 0]$ . Then a value of $α$ is

Select Answer:

Visualized Solution

Identifying the Matrix $A$

Geometric Effect of $A$

The Power Property of Rotation Matrices

Applying the Power $n = 32$

Analyzing the Target Matrix

Equating the Matrices

Solving for the Total Angle

Final Calculation for $α$

The Sigma Insight: Algebraic Operations on Matrices

The Geometry of Matrices

The Power of Iteration

Bridging Algebra and Geometry

The Final Resolution

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