Question

If \(A=\begin{bmatrix}\cos\theta & i\sin\theta\\ i\sin\theta & \cos\theta\end{bmatrix},\ \left(\theta=\frac{\pi}{24}\right)\) and \(A^5=\begin{bmatrix}a & b\\ c & d\end{bmatrix}\), where \(i=\sqrt{-1}\), then which one of the following is not true?

• A. \(a^2-d^2=0\)
• B. \(a^2-b^2=\frac{1}{2}\)
• C. \(0\le a^2+b^2\le 1\)
• D. \(a^2-c^2=1\)

Hint:

Show by induction that \(A^n=\begin{bmatrix}\cos n\theta & i\sin n\theta\\ i\sin n\theta & \cos n\theta\end{bmatrix}\), so \(a=d=\cos5\theta\), \(b=c=i\sin5\theta\).

Answer 1

The Correct Option is B

\(A^2=\begin{bmatrix}\cos^2\theta-\sin^2\theta & 2i\sin\theta\cos\theta\\ 2i\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta\end{bmatrix}=\begin{bmatrix}\cos2\theta & i\sin2\theta\\ i\sin2\theta & \cos2\theta\end{bmatrix}\); by induction \(A^n=\begin{bmatrix}\cos n\theta & i\sin n\theta\\ i\sin n\theta & \cos n\theta\end{bmatrix}\).
So \(a=d=\cos\frac{5\pi}{24}\) and \(b=c=i\sin\frac{5\pi}{24}\).
\(a^2-d^2=0\) (true). \(a^2-c^2=\cos^2+\sin^2=1\) (true). \(a^2-b^2=\cos^2+\sin^2=1\neq\frac12\) (false).
\(a^2+b^2=\cos^2\frac{5\pi}{24}-\sin^2\frac{5\pi}{24}=\cos\frac{5\pi}{12}\approx0.26\in[0,1]\) (true).
The statement that is not true is \(a^2-b^2=\frac12\).
Answer: (2) \(a^2-b^2=\frac{1}{2}\)