Question

If \( X = \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} \) and \( Y = \begin{bmatrix} 0 & 1 -1 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} \cos \theta & \sin \theta -\sin \theta & \cos \theta \end{bmatrix} \), then \( B \) equals:

• A. \( -X \cos \theta + Y \sin \theta \)
• B. \( X \sin \theta + Y \cos \theta \)
• C. \( X \cos \theta - Y \sin \theta \)
• D. \( X \cos \theta + Y \sin \theta \)

Hint:

Identity matrix \( X \) scalar multiplication preserves diagonal structure, while matrix \( Y \) introduces rotation-type structureCombine both to reconstruct rotation matrices easily.

Answer 1

The Correct Option is D

Step 1: Identify matrices.
Given:
\[ X = \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix}, \quad Y = \begin{bmatrix} 0 & 1 -1 & 0 \end{bmatrix} \]

Step 2: Multiply \( X \cos \theta \).
\[ X \cos \theta = \cos \theta \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} = \begin{bmatrix} \cos \theta & 0 0 & \cos \theta \end{bmatrix} \]

Step 3: Multiply \( Y \sin \theta \).
\[ Y \sin \theta = \sin \theta \begin{bmatrix} 0 & 1 -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \sin \theta -\sin \theta & 0 \end{bmatrix} \]

Step 4: Add both matrices.
\[ X \cos \theta + Y \sin \theta = \begin{bmatrix} \cos \theta & 0 0 & \cos \theta \end{bmatrix} + \begin{bmatrix} 0 & \sin \theta -\sin \theta & 0 \end{bmatrix} \]
\[ = \begin{bmatrix} \cos \theta & \sin \theta -\sin \theta & \cos \theta \end{bmatrix} \]

Step 5: Compare with given matrix \( B \).
\[ B = \begin{bmatrix} \cos \theta & \sin \theta -\sin \theta & \cos \theta \end{bmatrix} \]
Clearly, both matrices are identical.

Step 6: Final conclusion.
\[ \boxed{B = X \cos \theta + Y \sin \theta} \]