Question

If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \) and \( [F(x)]^2 = F(kx) \), then the value of \( k \) is:

• A. \( 1 \)
• B. \( 2 \)
• C. \( 0 \)
• D. \( -2 \)

Hint:

When comparing matrix equations involving trigonometric functions, focus on the diagonal and off-diagonal elements to identify patterns or relationships. This simplifies finding constants like \( k \).

Answer 1

The Correct Option is B

Step 1: Calculate \( [F(x)]^2 \).
The given matrix \( F(x) \) is: \[ F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] To find \( [F(x)]^2 \), perform the matrix multiplication: \[ [F(x)]^2 = F(x) \cdot F(x). \] Carrying out the multiplication: \[ \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos(2x) & -\sin(2x) & 0 \\ \sin(2x) & \cos(2x) & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] Step 2: Relate \( [F(x)]^2 \) to \( F(kx) \).
The matrix \( F(kx) \) is given as: \[ F(kx) = \begin{bmatrix} \cos(kx) & -\sin(kx) & 0 \\ \sin(kx) & \cos(kx) & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] From the condition \( [F(x)]^2 = F(kx) \), compare the corresponding entries: \[ \cos(2x) = \cos(kx) \quad \text{and} \quad \sin(2x) = \sin(kx). \] Using the periodicity and uniqueness of trigonometric functions, we find: \[ kx = 2x \quad \Rightarrow \quad k = 2. \] Step 3: Final result.
The value of \( k \) is: \[ \boxed{2}. \]