Question

Consider the matrix \( f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
Given below are two statements:
Statement I: \( f(-x) \) is the inverse of the matrix \( f(x) \).
Statement II: \( f(x) f(y) = f(x + y) \).
In the light of the above statements, choose the correct answer from the options given below:"

• A. Statement I is false but Statement II is true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Both Statement I and Statement II are true

Answer 1

The Correct Option is D

To determine the truth of the given statements regarding the matrix \( f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \), we'll analyze each statement individually. 

1. Statement I: \( f(-x) \) is the inverse of the matrix \( f(x) \).
   • The inverse of a rotation matrix is its transpose when dealing with orthogonal matrices. For \( f(x) \), which is a rotation matrix in 3D space, the transpose should be equal to \( f(-x) \).
   • We calculate \( f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
   • Next, find the transpose of \( f(x) \): \(f(x)^T = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
   • As \( f(-x) = f(x)^T \), \( f(-x) \) indeed serves as the inverse of \( f(x) \).
2. Statement II: \( f(x) f(y) = f(x + y) \).
   • Recall that matrix multiplication of rotation matrices corresponds to the addition of angles. We calculate.
   • \( f(x) f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \).
   • Computing the product gives: \(f(x)f(y) = \begin{bmatrix} \cos x \cos y - \sin x \sin y & -(\cos x \sin y + \sin x \cos y) & 0 \\ \sin x \cos y + \cos x \sin y & \cos x \cos y - \sin x \sin y & 0 \\ 0 & 0 & 1 \end{bmatrix}\).
   • This matches with \(f(x+y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}\) after using trigonometric angle addition identities:
     • \(\cos(x+y) = \cos x \cos y - \sin x \sin y\)
     • \(\sin(x+y) = \sin x \cos y + \cos x \sin y\)

Since both statements hold true, the correct answer is: Both Statement I and Statement II are true.

Answer 2

Step 1. Verification of Statement I: To check if \( f(-x) \) is the inverse of \( f(x) \), we need to verify if \( f(x) \cdot f(-x) = I \), where \( I \) is the identity matrix.

  - Calculate \( f(-x) \):
 
    \(f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
   
  
\(f(-x) = \begin{bmatrix} \cos(-x) & -\sin(-x) & 0 \\ \sin(-x) & \cos(-x) & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)

  - Now, compute \( f(x) \cdot f(-x) \):
\(f(x) \cdot f(-x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I\)

  - Thus, \( f(-x) \) is indeed the inverse of \( f(x) \), so Statement I is true.

Step 2. Verification of Statement II: To verify \( f(x) \cdot f(y) = f(x + y) \), perform the matrix multiplication \( f(x) \cdot f(y) \):

 \(f(x) \cdot f(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(x + y) & -\sin(x + y) & 0 \\ \sin(x + y) & \cos(x + y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = f(x + y)\)
 

  - Therefore, \( f(x) \cdot f(y) = f(x + y) \), so Statement II is also true.

Since both Statement I and Statement II are true, the correct answer is \( (4) \).