Question:
If \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), verify that \( A'A=I \).
If \( A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \), verify that \( A'A=I \).
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The matrix given in the question is a rotation matrix. A key property of rotation matrices is that they are orthogonal, which means their transpose is equal to their inverse (\(A' = A^{-1}\)). Therefore, \( A'A = A^{-1}A = I \). Recognizing this can give you confidence in your result before you even start the calculation.
Updated On: Sep 6, 2025
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Solution and Explanation
Step 1: Understanding the Concept:
This question requires verifying a property of a given matrix. It involves finding the transpose of the matrix \( A \) (denoted as \( A' \)) and then performing matrix multiplication of \( A' \) with \( A \). The result should be the identity matrix \( I \). Such a matrix \( A \) is called an orthogonal matrix.
Step 2: Key Formula or Approach:
1. Find the transpose of matrix \( A \), denoted as \( A' \).
2. Multiply \( A' \) by \( A \).
3. Use the trigonometric identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \).
4. The identity matrix of order 2 is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Step 3: Detailed Explanation:
The given matrix is: \[ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] First, we find the transpose of \( A \), \( A' \): \[ A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \] Next, we compute the product \( A'A \): \[ A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] Performing the multiplication: \[ A'A = \begin{bmatrix} (\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) \\ (\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) \end{bmatrix} \] \[ A'A = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \] Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \): \[ A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] This is the identity matrix \( I \).
Step 4: Final Answer:
Since \( A'A = I \), the property is verified. ✅
This question requires verifying a property of a given matrix. It involves finding the transpose of the matrix \( A \) (denoted as \( A' \)) and then performing matrix multiplication of \( A' \) with \( A \). The result should be the identity matrix \( I \). Such a matrix \( A \) is called an orthogonal matrix.
Step 2: Key Formula or Approach:
1. Find the transpose of matrix \( A \), denoted as \( A' \).
2. Multiply \( A' \) by \( A \).
3. Use the trigonometric identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \).
4. The identity matrix of order 2 is \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Step 3: Detailed Explanation:
The given matrix is: \[ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] First, we find the transpose of \( A \), \( A' \): \[ A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \] Next, we compute the product \( A'A \): \[ A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \] Performing the multiplication: \[ A'A = \begin{bmatrix} (\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) & (\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) \\ (\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) & (\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) \end{bmatrix} \] \[ A'A = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & \cos \alpha \sin \alpha - \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha - \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} \] Using the identity \( \cos^2 \alpha + \sin^2 \alpha = 1 \): \[ A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] This is the identity matrix \( I \).
Step 4: Final Answer:
Since \( A'A = I \), the property is verified. ✅
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