Question

Let \(\theta=\frac{\pi}{5}\) and \(A=\begin{bmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{bmatrix}\). If \(B=A+A^4\), then det (B) :

• A. lies in (1, 2)
• B. lies in (2, 3).
• C. is zero.
• D. is one.

Hint:

A is a rotation-type matrix, so \(A^n\) has \(n\theta\) in place of \(\theta\).

Answer 1

The Correct Option is A

For \(A=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\) we have \(A^n=\begin{bmatrix}\cos n\theta&\sin n\theta\\-\sin n\theta&\cos n\theta\end{bmatrix}\).
\(B=A+A^4=\begin{bmatrix}\cos\theta+\cos4\theta&\sin\theta+\sin4\theta\\-(\sin\theta+\sin4\theta)&\cos\theta+\cos4\theta\end{bmatrix}\).
\(\det B=(\cos\theta+\cos4\theta)^2+(\sin\theta+\sin4\theta)^2=2+2\cos3\theta\).
With \(\theta=\frac{\pi}{5}\): \(\det B=2+2\cos\frac{3\pi}{5}=2-2\sin\frac{\pi}{10}=2-2\cdot0.309\approx1.38\).
This lies in (1, 2).
Answer: (1) lies in (1, 2)