If \( A=\begin{bmatrix}0 & -\tan\alpha/2\\ \tan\alpha/2 & 0\end{bmatrix} \), then the value of \( \begin{bmatrix}\cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha\end{bmatrix} \) for all \( \alpha\in\mathrm{R} \) (set of real numbers), is
(I is identity matrix of order 2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (I+A)(I-A)^{-1} \)
Let \( t=\tan\frac{\alpha}{2} \), so \( A=\begin{bmatrix}0&-t\\t&0\end{bmatrix} \). Then
\( \displaystyle I+A=\begin{bmatrix}1&-t\\t&1\end{bmatrix} \)
\( \displaystyle I-A=\begin{bmatrix}1&t\\-t&1\end{bmatrix} \)
The determinant of \( I-A \) is \( 1+t^{2} \), so
\( \displaystyle (I-A)^{-1}=\frac{1}{1+t^{2}}\begin{bmatrix}1&-t\\t&1\end{bmatrix} \)
Multiplying,
\( \displaystyle (I+A)(I-A)^{-1}=\frac{1}{1+t^{2}}\begin{bmatrix}1&-t\\t&1\end{bmatrix}\begin{bmatrix}1&-t\\t&1\end{bmatrix} \)
\( \displaystyle =\frac{1}{1+t^{2}}\begin{bmatrix}1-t^{2}&-2t\\2t&1-t^{2}\end{bmatrix} \)
Using the half-angle forms \( \cos\alpha=\frac{1-t^{2}}{1+t^{2}} \) and \( \sin\alpha=\frac{2t}{1+t^{2}} \), this is
\( \displaystyle \begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix} \)
Why the other options are incorrect
Option (1) \( I \) — Incorrect. The identity matrix equals the given matrix only when \( \alpha=0 \).
Option (3) \( (I-A)(I+A)^{-1} \) — Incorrect. This is the rotation through −α; it has \( \sin\alpha \) and \( -\sin\alpha \) interchanged.
Option (4) \( A^{2} \) — Incorrect. \( A^{2}=-t^{2}I \) is a scalar matrix, not a rotation matrix.