Key Concept: Rotation Matrices and their Properties

The given matrix $A$ is a special type of matrix known as a rotation matrix. A 2D rotation matrix $R(\theta)$ rotates a vector in the plane counter-clockwise by an angle $\theta$. Its standard form is: $R(\theta) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ A fundamental property of rotation matrices is that multiplying two rotation matrices results in a rotation by the sum of their individual angles: $R(\theta_1) R(\theta_2) = R(\theta_1 + \theta_2)$ From this, it logically follows that raising a rotation matrix to an integer power $n$ is equivalent to rotating by $n$ times the original angle: $R(\theta)^n = R(n\theta)$ This property is extremely powerful for simplifying matrix powers that would otherwise involve many tedious matrix multiplications.

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Step-by-Step Solution

Step 1: Identify the given matrix $A$ as a rotation matrix. The matrix $A$ is given as: $A = \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}$ By comparing this with the general form of a rotation matrix $R(\theta)$, we can clearly see that $A$ is a rotation matrix corresponding to a rotation by an angle $\alpha$. So, we can write $A = R(\alpha)$.

• Why this step? Recognizing $A$ as a rotation matrix is the crucial first step. It allows us to apply the unique and simplifying properties of rotation matrices to calculate its powers efficiently, rather than performing repeated matrix multiplications.

Step 2: Calculate $A^{32}$ using the property of rotation matrices. Since $A = R(\alpha)$, we can use the property $R(\theta)^n = R(n\theta)$ to find $A^{32}$: $A^{32} = (R(\alpha))^{32} = R(32\alpha)$ Therefore, the matrix $A^{32}$ can be written as: $A^{32} = \begin{pmatrix} \cos(32\alpha) & -\sin(32\alpha) \\ \sin(32\alpha) & \cos(32\alpha) \end{pmatrix}$

• Why this step? This step provides an elegant and efficient way to compute the $32^{nd}$ power of matrix $A$. Instead of 31 complex matrix multiplications, which are prone to errors and time-consuming, we simply multiply the angle $\alpha$ by the power $32$. This leverages the geometric interpretation of the matrix.

Step 3: Equate the calculated $A^{32}$ with the given target matrix. The problem states that $A^{32}$ must be equal to the matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Equating our derived expression for $A^{32}$ with the given matrix: $\begin{pmatrix} \cos(32\alpha) & -\sin(32\alpha) \\ \sin(32\alpha) & \cos(32\alpha) \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

• Why this step? This step translates the problem's condition into a direct matrix equality. This equality is the basis for forming the trigonometric equations necessary to solve for $\alpha$.

Step 4: Formulate a system of trigonometric equations. For two matrices to be equal, their corresponding elements must be equal. By comparing the elements of the matrices from Step 3, we obtain a system of two trigonometric equations:

1. $\cos(32\alpha) = 0$
2. $\sin(32\alpha) = 1$

• Why this step? This step converts the matrix equation into a more familiar system of scalar trigonometric equations. Solving these equations simultaneously will give us the specific values for $32\alpha$ that satisfy the problem's conditions. Both equations must be satisfied for the matrix equality to hold true.

Step 5: Solve the system of trigonometric equations for $32\alpha$. We need to find the values of an angle, let's call it $\theta = 32\alpha$, such that $\cos \theta = 0$ AND $\sin \theta = 1$.

• The general solution for $\cos \theta = 0$ is $\theta = (2n+1)\frac{\pi}{2}$, where $n \in \mathbb{Z}$. This means $\theta$ can be $\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots$ or $-\frac{\pi}{2}, -\frac{3\pi}{2}, \dots$.
• The general solution for $\sin \theta = 1$ is $\theta = 2k\pi + \frac{\pi}{2}$, where $k \in \mathbb{Z}$. This means $\theta$ can be $\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, \dots$.

For both conditions to be satisfied simultaneously, $\theta$ must be of the form $2k\pi + \frac{\pi}{2}$. Thus, we set: $32\alpha = 2k\pi + \frac{\pi}{2} \quad \text{for some integer } k$

• Why this step? This step identifies the precise angle(s) on the unit circle where the cosine is zero and the sine is one. This corresponds to the positive y-axis. The general solution ensures we capture all possible values for $32\alpha$ that satisfy the conditions.

Step 6: Solve for $\alpha$ and select a value from the options. Now, we isolate $\alpha$ from the equation derived in Step 5: $32\alpha = 2k\pi + \frac{\pi}{2}$ Divide both sides by 32: $\alpha = \frac{1}{32} \left( 2k\pi + \frac{\pi}{2} \right)$ Distribute the $\frac{1}{32}$: $\alpha = \frac{2k\pi}{32} + \frac{\pi}{64}$ Simplify the first term: $\alpha = \frac{k\pi}{16} + \frac{\pi}{64}$ We need to find a value of $\alpha$ that matches one of the given options. Let's test integer values for $k$:

• For $k=0$: $\alpha = \frac{0 \cdot \pi}{16} + \frac{\pi}{64} = \frac{\pi}{64}$.
• For $k=1$: $\alpha = \frac{\pi}{16} + \frac{\pi}{64} = \frac{4\pi}{64} + \frac{\pi}{64} = \frac{5\pi}{64}$.
• For $k=-1$: $\alpha = -\frac{\pi}{16} + \frac{\pi}{64} = -\frac{4\pi}{64} + \frac{\pi}{64} = -\frac{3\pi}{64}$.

Comparing these values with the given options: (A) 0 (B) ${\pi \over {16}}$ (C) ${\pi \over {32}}$ (D) ${\pi \over {64}}$

We find that for $k=0$, $\alpha = \frac{\pi}{64}$, which matches option (D).

• Why this step? This is the final step where we determine the specific value of $\alpha$ that satisfies the initial matrix equation. We derive the general solution for $\alpha$ and then check which specific value from the options fits this general solution.

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Tips for Success & Common Mistakes

• Recognize Standard Forms: Always be on the lookout for standard matrix forms like rotation matrices, identity matrices, or scalar matrices. Recognizing them early can save a lot of computation.
• General Solutions for Trigonometric Equations: When solving trigonometric equations, remember to use general solutions (e.g., $2k\pi + \theta_0$) to cover all possibilities.
• Simultaneous Conditions: If multiple trigonometric conditions arise from matrix equality, ensure that the angle satisfies ALL conditions simultaneously.
• Careful with Arithmetic: Even simple division or fraction addition can lead to errors. Double-check your calculations.
• Options Check: After finding the general solution for $\alpha$, substitute values of $k$ (starting with $k=0, 1, -1$) to see if any of the options are generated.

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Summary & Key Takeaway

This problem beautifully demonstrates the power of recognizing special matrix types, specifically rotation matrices. By understanding their properties, we can efficiently calculate high powers of such matrices without performing laborious multiplications. The problem then reduces to solving a system of basic trigonometric equations. The key takeaway is that leveraging matrix properties and their geometric interpretations can significantly simplify complex-looking problems in matrices.