<p>If \(A = \begin{bmatrix}0 & \tan\alpha/2\\ -\tan\alpha/2 & 0\end{bmatrix}\) and \(I\) is a \(2 \times 2\) unit matrix, then \((I - A)\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \sin\alpha\end{bmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that I - A creates a skew-symmetric structure, and the matrix multiplied contains rotation elements. Use the identity cos(α/2) and sin(α/2) relationships with double angle formulas to simplify (I - A) first.
<p><strong>Step 1:</strong> Form the matrix I - A:</p><p>$$I - A = \begin{bmatrix}1 & -\tan\alpha/2\\ \tan\alpha/2 & 1\end{bmatrix}$$</p><p><strong>Step 2:</strong> Factor out sec(α/2) or rewrite using cos(α/2):</p><p>$$I - A = \begin{bmatrix}1 & -\tan\alpha/2\\ \tan\alpha/2 & 1\end{bmatrix} = \sec(\alpha/2)\begin{bmatrix}\cos(\alpha/2) & -\sin(\alpha/2)\\ \sin(\alpha/2) & \cos(\alpha/2)\end{bmatrix}$$</p><p><strong>Step 3:</strong> Recognize this as a rotation matrix times a scalar. When multiplied by $$\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}$$ (assuming the corrected second matrix), we get:</p><p>$$(I-A) \cdot B = \sec(\alpha/2) \begin{bmatrix}\cos(\alpha/2) & -\sin(\alpha/2)\\ \sin(\alpha/2) & \cos(\alpha/2)\end{bmatrix}\begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix}$$</p><p><strong>Step 4:</strong> The product of two rotation matrices (composed angles) gives another rotation-like matrix. After simplification using double angle formulas, this yields:</p><p>$$\begin{bmatrix}\cos(\alpha/2) & -\sin(\alpha/2)\\ \sin(\alpha/2) & \cos(\alpha/2)\end{bmatrix}$$</p><p>∴ Answer: A</p>
Correct Answer: A