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 ____.
Step-by-Step Solution
Key Concept: Use the property that for a matrix A of the form [[0, -t], [t, 0]], (I+A)(I-A)^-1 is a rotation matrix. Specifically, let t = tan(theta/2). Then (I+A)(I-A)^-1 = [[cos(theta), -sin(theta)], [sin(theta), cos(theta)]]. Thus a = cos(theta) and b = sin(theta). Then a^2 + b^2 = 1.
Let $t = \tan(\theta/2)$. Then $A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix}$. $I+A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$. $I-A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}$. $(I-A)^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}$. $(I+A)(I-A)^{-1} = \frac{1}{1+t^2} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = \frac{1}{1+t^2} \begin{bmatrix} 1-t^2 & -2t \\ 2t & 1-t^2 \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$. Thus $a = \cos\theta$ and $b = \sin\theta$. $a^2+b^2 = \cos^2\theta + \sin^2\theta = 1$. Therefore $13(a^2+b^2) = 13(1) = 13$.
Correct Answer: 1