Here's a detailed and educational solution to the problem:

Understanding the Key Concepts

The problem involves a special type of matrix known as a rotation matrix. A $2 \times 2$ matrix of the form R(\theta) = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right] represents a counter-clockwise rotation by an angle $\theta$ in a 2D plane.

A crucial property of rotation matrices is that if $A = R(\theta)$, then $A^n = R(n\theta)$. This means applying a rotation by $\theta$ n times is equivalent to a single rotation by $n\theta$.

We will also use the determinant formula for a $2 \times 2$ matrix: for $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det(M) = ad - bc$. Finally, several trigonometric identities will be essential:

• $\cos^2 x + \sin^2 x = 1$
• $\cos A \cos B + \sin A \sin B = \cos(A-B)$
• $\cos(\pi - x) = -\cos x$
• $\cos(90^\circ + x) = -\sin x$
• The standard value of $\sin(18^\circ) = \frac{\sqrt{5}-1}{4}$.

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Step-by-Step Solution

Step 1: Identify Matrix A as a Rotation Matrix and find its general power $A^n$.

The given matrix is: A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right] This matrix is precisely a rotation matrix, $R(\theta)$, representing a rotation by an angle $\theta$.

Let's compute $A^2$ to see the pattern: A^2 = A \cdot A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right] \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right] A^2 = \left[ {\matrix{ {\cos^2 \theta - \sin^2 \theta } & {\cos \theta \sin \theta + \sin \theta \cos \theta } \cr { - \sin \theta \cos \theta - \cos \theta \sin \theta } & { - \sin^2 \theta + \cos^2 \theta } \cr } } \right] Using the double angle formulas ($\cos 2\theta = \cos^2 \theta - \sin^2 \theta$ and $\sin 2\theta = 2\sin \theta \cos \theta$): A^2 = \left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right] This confirms the property for $n=2$. By repeatedly applying this multiplication (or more formally, by mathematical induction), we can generalize this to any positive integer $n$: A^n = \left[ {\matrix{ {\cos n\theta } & {\sin n\theta } \cr { - \sin n\theta } & {\cos n\theta } \cr } } \right] Tip: Recognizing this specific matrix form as a rotation matrix is key to solving many problems efficiently. If you don't recognize it, you'd have to perform matrix multiplication repeatedly, which is prone to errors and time-consuming.

Step 2: Calculate $A^4$.

Using the general formula for $A^n$ with $n=4$: A^4 = \left[ {\matrix{ {\cos 4\theta } & {\sin 4\theta } \cr { - \sin 4\theta } & {\cos 4\theta } \cr } } \right]

Step 3: Calculate Matrix B.

The problem defines $B = A + A^4$. We perform element-wise matrix addition: B = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right] + \left[ {\matrix{ {\cos 4\theta } & {\sin 4\theta } \cr { - \sin 4\theta } & {\cos 4\theta } \cr } } \right] B = \left[ {\matrix{ {\cos \theta + \cos 4\theta } & {\sin \theta + \sin 4\theta } \cr { - \sin \theta - \sin 4\theta } & {\cos \theta + \cos 4\theta } \cr } } \right] B = \left[ {\matrix{ {\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \cr { - (\sin 4\theta + \sin \theta) } & {\cos 4\theta + \cos \theta } \cr } } \right]

Step 4: Calculate $\det(B)$.

For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det(M) = ad - bc$. Here, $a = \cos 4\theta + \cos \theta$, $b = \sin 4\theta + \sin \theta$, $c = -(\sin 4\theta + \sin \theta)$, and $d = \cos 4\theta + \cos \theta$. $\det(B) = (\cos 4\theta + \cos \theta)(\cos 4\theta + \cos \theta) - (\sin 4\theta + \sin \theta)(-(\sin 4\theta + \sin \theta))$ $\det(B) = (\cos 4\theta + \cos \theta)^2 + (\sin 4\theta + \sin \theta)^2$ Expand the squares: $\det(B) = (\cos^2 4\theta + \cos^2 \theta + 2\cos 4\theta \cos \theta) + (\sin^2 4\theta + \sin^2 \theta + 2\sin 4\theta \sin \theta)$ Rearrange terms and group them to use trigonometric identities: $\det(B) = (\cos^2 4\theta + \sin^2 4\theta) + (\cos^2 \theta + \sin^2 \theta) + 2(\cos 4\theta \cos \theta + \sin 4\theta \sin \theta)$ Using the identity $\cos^2 x + \sin^2 x = 1$: $\det(B) = 1 + 1 + 2(\cos 4\theta \cos \theta + \sin 4\theta \sin \theta)$ Using the compound angle identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $\det(B) = 2 + 2 \cos(4\theta - \theta)$ $\det(B) = 2 + 2 \cos(3\theta)$ Tip: Be careful with signs when applying the determinant formula, especially when elements are negative. Also, always look for opportunities to simplify using fundamental trigonometric identities.

Step 5: Substitute the value of $\theta$ and evaluate the trigonometric term.

Given $\theta = \frac{\pi}{5}$. Substitute this into the expression for $\det(B)$: $\det(B) = 2 + 2 \cos\left(3 \cdot \frac{\pi}{5}\right) = 2 + 2 \cos\left(\frac{3\pi}{5}\right)$ Now, we need to evaluate $\cos\left(\frac{3\pi}{5}\right)$. We know that $\frac{3\pi}{5}$ is $108^\circ$. We can use the identity $\cos(\pi - x) = -\cos x$: $\cos\left(\frac{3\pi}{5}\right) = \cos\left(\pi - \frac{2\pi}{5}\right) = -\cos\left(\frac{2\pi}{5}\right)$ Alternatively, we can use $\cos(90^\circ + x) = -\sin x$: $\cos\left(\frac{3\pi}{5}\right) = \cos(108^\circ) = \cos(90^\circ + 18^\circ) = -\sin(18^\circ)$ The value of $\sin(18^\circ)$ is a standard result that should be memorized or derived: $\sin(18^\circ) = \frac{\sqrt{5}-1}{4}$ Therefore, $\cos\left(\frac{3\pi}{5}\right) = -\frac{\sqrt{5}-1}{4}$ Substitute this back into the determinant expression: $\det(B) = 2 + 2 \left(-\frac{\sqrt{5}-1}{4}\right)$ $\det(B) = 2 - \frac{\sqrt{5}-1}{2}$ To simplify, find a common denominator: $\det(B) = \frac{4 - (\sqrt{5}-1)}{2} = \frac{4 - \sqrt{5} + 1}{2} = \frac{5 - \sqrt{5}}{2}$ **Common Mistake