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:

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For matrix equations involving trigonometric transformations, compare individual elements to find relationships.
Updated On: Jan 28, 2025
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The Correct Option is B

Solution and Explanation

Step 1: Compute \( [F(x)]^2 \). 
The given matrix \( F(x) \) is: \[ F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] Using matrix multiplication: \[ [F(x)]^2 = F(x) \cdot F(x). \] Perform the multiplication: \[ \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] The resulting matrix is: \[ [F(x)]^2 = \begin{bmatrix} \cos(2x) & -\sin(2x) & 0 \\ \sin(2x) & \cos(2x) & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] Step 2: Compare with \( F(kx) \). 
The matrix \( F(kx) \) is: \[ F(kx) = \begin{bmatrix} \cos(kx) & -\sin(kx) & 0 \\ \sin(kx) & \cos(kx) & 0 \\ 0 & 0 & 1 \end{bmatrix}. \] From the condition \( [F(x)]^2 = F(kx) \), we compare: \[ \cos(2x) = \cos(kx) \quad \text{and} \quad \sin(2x) = \sin(kx). \] This implies \( kx = 2x \), so \( k = 2 \). 
Step 3: Conclusion. 
The value of \( k \) is: \[ \boxed{2}. \]

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