1. Identify the matrix form

Given

$$A=\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix}, \qquad \theta=\frac{\pi}{5}$$

This is a rotation-type matrix. For such a matrix,

$$A^n=\begin{bmatrix} \cos n\theta & \sin n\theta\\ -\sin n\theta & \cos n\theta \end{bmatrix}$$

So,

$$A^4=\begin{bmatrix} \cos 4\theta & \sin 4\theta\\ -\sin 4\theta & \cos 4\theta \end{bmatrix}$$

2. Compute $B=A+A^4$

Thus,

$$B=\begin{bmatrix} \cos\theta+\cos4\theta & \sin\theta+\sin4\theta\\ -(\sin\theta+\sin4\theta) & \cos\theta+\cos4\theta \end{bmatrix}$$

Let

$$a=\cos\theta+\cos4\theta, \qquad b=\sin\theta+\sin4\theta$$

Then

$$B=\begin{bmatrix}a & b\\ -b & a\end{bmatrix}$$

Hence,

$$\det(B)=a^2+b^2$$

3. Simplify using trigonometric identities

We use

$$(\cos\theta+\cos4\theta)^2+(\sin\theta+\sin4\theta)^2$$

Expand:

$$=\cos^2\theta+\cos^24\theta+2\cos\theta\cos4\theta+\sin^2\theta+\sin^24\theta+2\sin\theta\sin4\theta$$

Using $\cos^2 x+\sin^2 x=1$,

$$\det(B)=1+1+2(\cos\theta\cos4\theta+\sin\theta\sin4\theta)$$

Now,

$$\cos\theta\cos4\theta+\sin\theta\sin4\theta=\cos(\theta-4\theta)=\cos(-3\theta)=\cos3\theta$$

Therefore,

$$\det(B)=2+2\cos3\theta$$

4. Substitute $\theta=\frac{\pi}{5}$

Then

$$3\theta=\frac{3\pi}{5}$$

So,

$$\det(B)=2+2\cos\frac{3\pi}{5}$$

Now,

$$\cos\frac{3\pi}{5}=\cos108^\circ=-\cos72^\circ$$

And

$$\cos72^\circ=\frac{\sqrt5-1}{4}$$

Hence,

$$\cos\frac{3\pi}{5}=-\frac{\sqrt5-1}{4}=\frac{1-\sqrt5}{4}$$

So,

$$\det(B)=2+2\cdot\frac{1-\sqrt5}{4} =2+\frac{1-\sqrt5}{2} =\frac{5-\sqrt5}{2}$$

5. Check the interval

Using $\sqrt5\approx 2.236$,

$$\det(B)=\frac{5-2.236}{2}=\frac{2.764}{2}\approx 1.382$$

Thus,

$$\det(B)\in(1,2)$$

6. Evaluate options

• A: lies in $(1,2)$ ✅
• B: lies in $(2,3)$ ❌
• C: is zero ❌
• D: is one ❌

Therefore, the correct option is A.