Problem 56 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\mathrm{A}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right]\), then the matrix \(\mathrm{A}^{-50}\) when \(\theta=\frac{\pi}{12}\), is equal to: \(\quad\) Jan 09, 2019 (I)] (a) \(\left[\begin{array}{cc}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]\) (b) \(\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]\) (c) \(\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]\) (d) \(\left[\begin{array}{cc}\frac{1}{2} & \frac{\sqrt{3}}{2} \\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{array}\right]\)

Step-by-Step Solution

Answer

The matrix \( A^{-50} \) is equal to option (c): \( \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \).

Step 1: Understanding the problem

We are given a rotation matrix \( A \) for a rotation by angle \( \theta \), and we need to find \( A^{-50} \). The matrix \( A \) is a standard rotation matrix in the 2D plane.

Step 2: Matrix components for given \( \theta \)

First, substitute \( \theta = \frac{\pi}{12} \) into the rotation matrix components to find \( \cos \theta \) and \( \sin \theta \): - \( \cos \left(\frac{\pi}{12}\right) = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)- \( \sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \).However, these values are expected to eventually simplify in the matrix powers.

Step 3: Properties of rotation matrices

Note that for a 2D rotation matrix like \( A \), \( A^{-1} = A^T \) (the transpose). Hence, \( A^{-50} = (A^{-1})^{50} = (A^T)^{50} \). The matrix transpose of a rotation matrix changes its angle sign.

Step 4: Computing \( A^{50} \) from \( A^{-50} \) via angles

Instead of directly calculating the powers, consider angle based property: - A matrix \( A^n \) represents a rotation by \( n\theta \).- Thus, \( A^{-50} \) corresponds to a rotation by \(-50 \times \frac{\pi}{12}\).

Step 5: Simplifying the angle

Simplify \(-50 \times \frac{\pi}{12} = -\frac{50\pi}{12} = -\frac{25\pi}{6}\). Convert this angle into an equivalent angle within \([0, 2\pi)\): - Add \( 8\pi \) (equivalent to 4 full rotations):- \(-\frac{25\pi}{6} + \frac{48\pi}{6} = \frac{23\pi}{6}\).

Step 6: Identifying the equivalent angle

Since \( \frac{23\pi}{6} \) is larger than \( 2\pi \), find its equivalent angle by subtracting \( 2\pi \) until in \([0, 2\pi)\): - \( \frac{23\pi}{6} - \frac{12\pi}{6} = \frac{11\pi}{6} \).- Further reduce within the range \([0, 2\pi)\):- \( \frac{11\pi}{6} \) is equivalent to \( \pi - \frac{\pi}{6} = \frac{5\pi}{6} \equiv -\frac{\pi}{6} \) in terms of cosine/sine.

Step 7: Final rotation matrix

From the equivalent angle \(-\frac{\pi}{6}\), use standard trigonometric values:- \( \cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \)- \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \).Thus, the matrix is: \[\begin{bmatrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \-\frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}\]

Step 8: Compare with options

Compare the derived matrix with the given options: - The derived matrix matches option (c).

Key Concepts

Matrix Inversion

Matrix inversion is a concept where we find a matrix that, when multiplied by the original matrix, results in the identity matrix. This is equivalent to finding the "inverse" operation to what the original matrix performs.
Rotation matrices have a convenient property: in 2D transformations, the inverse of a rotation matrix is simply its transpose.

• A 2D rotation matrix, \( A \), is given by: \[ A = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \]
• The inverse is \( A^{-1} = A^T \), which means: \[ A^{-1} = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix} \]

This inverse matrix will rotate any vector that was originally rotated by \( A \) back to its initial position, thereby highlighting the powerful geometric nature of matrix inversion.

Trigonometric Functions

Trigonometric functions describe relationships within a right triangle and are fundamental in describing periodic phenomena such as waves. The two basic trigonometric functions involved in this problem are sine and cosine, which are essential in rotation matrices.
For an angle \( \theta \):

• \( \cos \theta \) gives the adjacent side over the hypotenuse in a right triangle.
• \( \sin \theta \) measures the opposite side over the hypotenuse.

These functions help define the rotation and transformation matrices in 2D.
For rotation matrices, using values from trigonometric functions like \( \cos \left(\frac{\pi}{12}\right) \) and \( \sin \left(\frac{\pi}{12}\right) \), we can build the rotation matrix as shown. Often the exact trigonometric values are simplified during calculations by noting the periodicity and symmetry properties of these functions.

Angle Simplification

Simplifying an angle is an essential skill, especially in contexts involving periodic operations like rotations. Angles can often be reduced by leveraging the periodic property of trigonometric functions, which repeat every \( 2\pi \).

• When calculating powers of a rotation matrix, you encounter expressions like \(-50 \times \frac{\pi}{12}\).
• After simplification, the angle might go beyond a full circle (i.e., greater than \( 2\pi \) radians), needing conversion into an equivalent angle within the \([0, 2\pi)\) range.

For example, simplifying an angle like \(-\frac{25\pi}{6}\) involves:

• First, converting it to an equivalent positive angle by adding multiples of \( 2\pi \), resulting in \(\frac{23\pi}{6}\).
• Then reducing \(\frac{23\pi}{6}\) further by subtracting \(2\pi\) increments to bring it into the \([0, 2\pi)\) range, resulting in \(-\frac{\pi}{6}\).

Following these steps ensures computations remain manageable, avoiding excessive angles that complicate calculations while retaining the correct geometric interpretations.