Question

If \( A = \begin{bmatrix} 0 & -\tan\left(\frac{\theta}{2}\right) \\ \tan\left(\frac{\theta}{2}\right) & 0 \end{bmatrix} \) and \( (I_2 + A)(I_2 - A)^{-1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \), then \( 13(a^2 + b^2) \) is equal to __________.

Hint:

The matrix $(I+A)(I-A)^{-1}$ where $A$ is a skew-symmetric matrix is known as the Cayley transform; it always results in an orthogonal matrix (where $a^2 + b^2 = 1$).

Answer 1

Correct Answer: 13

Step 1: Let \( t = \tan\left(\frac{\theta}{2}\right) \). Then \( I + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \) and \( I - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix} \).

Step 2: \( (I - A)^{-1} = \dfrac{1}{1 + t^2} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}. \)

Step 3: \[ (I + A)(I - A)^{-1} = \dfrac{1}{1 + t^2} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \dfrac{1}{1 + t^2} \begin{bmatrix} 1 - t^2 & -2t \\ 2t & 1 - t^2 \end{bmatrix}. \]
Step 4: Comparing with \( \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \), we get \( a = \dfrac{1 - t^2}{1 + t^2} = \cos \theta \) and \( b = \dfrac{2t}{1 + t^2} = \sin \theta \).

Step 5: \( a^2 + b^2 = \cos^2 \theta + \sin^2 \theta = 1. \)

Step 6: \( 13(a^2 + b^2) = 13(1) = 13. \)