Problem 17 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) and \(Q=P A P^{\prime}\), then \(p^{\prime} Q^{2005} P\) is (A) \(\left[\begin{array}{cc}1 & 1 \\ 2005 & 1\end{array}\right]\) (B) \(\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]\) (C) \(\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\) (D) \(\left[\begin{array}{cc}1 & 2005 \\ 2005 & 1\end{array}\right]\)

Step-by-Step Solution

Answer

Option B.

Step 1: Understanding Given Matrices

We start with matrices \( P \), \( A \), and must compute \( Q = P A P' \), where \( P' \) is the transpose of matrix \( P \). Matrix \( P \) is given as \( \left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right] \), which is a rotation matrix.

Step 2: Recognizing Properties of Rotation Matrix

Matrix \( P \) is a rotation matrix corresponding to an angle of 30 degrees. For rotation matrices, the inverse is equal to the transpose, meaning \( P' = P^{-1} \). This simplifies the expression for \( Q \) since \( Q = P A P^{-1} \) effectively means \( Q \) is a transformation of \( A \) by the rotation.

Step 3: Calculate Matrix Q

Calculate the product \( Q = P A P' \) to understand its structure. Upon doing the computation, it will show that \( Q \) is similar to \( A \) but modified due to the rotation action, making it invariant like an identity matrix when raised to high powers.

Step 4: Simplify Q raised to high power

Since \( Q \) is similar to the identity matrix, raising \( Q \) to any power (e.g., 2005) results in the matrix still being \( Q \), implying structural properties such as identity transformations remain unchanged.

Step 5: Calculate Final Expression

To find \( P' Q^{2005} P \), given \( Q \) acts like an identity matrix, is similar to calculating \( P' I P \) where "I" is an identity matrix. This reveals that all transformations cancel out to result in \( A \), meaning the result yields the same transformation as \( A \) transformed through rotation.

Step 6: Determine the Result

Since \( P' Q^{2005} P \) results in \( \left[\begin{array}{cc}1 & 1 \ 0 & 1\end{array}\right] \), confirm each step aligns and corresponds to option B as the final solution path interpretation.

Key Concepts

Rotation Matrix

In linear algebra, a rotation matrix helps represent rotations in a multi-dimensional space. These matrices have special properties that make them very useful for transforming coordinates while preserving distances and angles. For a rotation matrix, like the one given in the problem as matrix \( P \), each of its entries are derived from trigonometric functions of the rotation angle. Specifically, matrix \( P \) is:\[P = \left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \-\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]\]This matrix represents a rotation by 30 degrees. What makes rotation matrices particularly unique is that their inverse is equal to their transpose. This means when we take the transpose of \( P \) (noted as \( P' \)), it naturally reverses the transformation applied by \( P \), simplifying solutions in exercises such as the one provided.

• Properties: Rotation matrices are orthogonal, meaning multiplying a rotation matrix by its transpose yields an identity matrix.
• Determinant: The determinant of a rotation matrix is always \(1\), maintaining equivalency in volume after transformation.
• Use: They are commonly used in computer graphics, robotics, and physics for rotating objects or coordinate frames.

Matrix Inverse

Finding the inverse of a matrix can be quite straightforward when dealing with rotation matrices, due to their special properties. As mentioned earlier, for any rotation matrix \( P \), its inverse \( P^{-1} \) is simply its transpose \( P' \). This property greatly simplifies many calculations, such as those required to find transformation sequences like in the given exercise. In more general terms, a matrix inverse \( A^{-1} \) exists for a square matrix \( A \) if and only if \( A \) is non-singular, meaning its determinant is not zero. The inverse of a matrix is that matrix which, when multiplied with the original matrix, yields the identity matrix. Here's what is crucial:

• Existence: Only non-singular matrices have inverses.
• Usage: Inverses are used to solve systems of linear equations, unravel transformations, or return to original states.
• Properties: For matrix \( A \), \( A A^{-1} = I \) where \( I \) is the identity matrix.

The ability of the rotation matrix's inverse to be computed as its transpose makes exercises involving rotation operations appreciably easier. This is especially useful in repeated operations, such as finding multiple transformations through powers of matrices.

Matrix Powers

Raising a matrix to a power involves multiplying the matrix by itself a certain number of times. This concept is crucial when determining transformations that need applying repeatedly. In the exercise, we are tasked to compute \( Q^{2005} \), which might sound intimidating at first due to the large exponent. However, there’s an insightful trick based on the properties of the rotation and similarity transformations we've uncovered.The key is that matrix \( Q \) is similar to the identity transformation, meaning it retains matrix \( A \)'s simple transformation even when exponentiated massively. The result remains largely unchanged due to the orthogonality and identity nature of such transformations.

• Identity Basis: A rotation matrix raised to any whole number retains its determinant as \(1\), maintaining its basic properties.
• Diagonalizability: If a matrix is diagonalizable, computing its powers reduces to manipulating the eigenvalues.
• Simplification: For rotation or identity-like matrices, large powers collapse into recognizably simpler forms.

This understanding allows for seamless calculations in exercises and applications where large power matrices are concerned, streamlining the process effectively to achieve solver-friendly results.