The given matrix \( A \) represents a 2D rotation matrix. For \( A \), multiplying \( A \) repeatedly corresponds to successive rotations by \( \theta \). Specifically:
\[ A^k = \begin{pmatrix} \cos(k\theta) & -\sin(k\theta) \\ \sin(k\theta) & \cos(k\theta) \end{pmatrix}. \]
Here, \( A^{100} \) represents a rotation by \( 100\theta \):
\[ A^{100} = \begin{pmatrix} \cos(100\theta) & -\sin(100\theta) \\ \sin(100\theta) & \cos(100\theta) \end{pmatrix}. \]
Step 1: Simplify \( 100\theta \mod 2\pi \).
Since \( \theta = \frac{2\pi}{7} \), we calculate:
\[ 100\theta = 100 \cdot \frac{2\pi}{7} = \frac{200\pi}{7}. \]
The angle \( \frac{200\pi}{7} \) can be reduced modulo \( 2\pi \). Divide 200 by 7:
\[ 200 \div 7 = 28 \quad (\text{remainder } 4). \]
Thus:
\[ 100\theta = \frac{200\pi}{7} = 28 \cdot 2\pi + \frac{8\pi}{7}. \]
Modulo \( 2\pi \), this reduces to:
\[ 100\theta \equiv \frac{8\pi}{7} \pmod{2\pi}. \]
Step 2: Express \( \frac{8\pi}{7} \) in terms of \( \theta \).
Since \( \theta = \frac{2\pi}{7} \), we write:
\[ 100\theta \equiv 4\theta \pmod{2\pi}. \]
Step 3: Compute \( A^{100} \).
Using the formula for \( A^k \), we find:
\[ A^{100} = \begin{pmatrix} \cos(4\theta) & -\sin(4\theta) \\ \sin(4\theta) & \cos(4\theta) \end{pmatrix}. \]
Substitute \( 4\theta \) into the matrix form:
\[ A^{100} = \begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix}. \]
Conclusion: The value of \( A^{100} \) is:
\[ \begin{pmatrix} \cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{pmatrix}. \]
If \[ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}, \] find \( A^{-1} \).
Using \( A^{-1} \), solve the following system of equations:
\[ \begin{aligned} 2x - 3y + 5z &= 11 \quad \text{(1)} \\ 3x + 2y - 4z &= -5 \quad \text{(2)} \\ x + y - 2z &= -3 \quad \text{(3)} \end{aligned} \]