Solution: Let us write \[ A(\theta ) = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \\ \end{pmatrix}. \] Note that \begin{align*} & A\left( \theta _1 \right) A\left( \theta _2 \right) \\ = & \begin{pmatrix} \cos \theta_1 & \sin \theta_1 \\ -\sin \theta_1 & \cos \theta_1 \\ \end{pmatrix} \begin{pmatrix} \cos \theta_2 & \sin \theta_2 \\ -\sin \theta_2 & \cos \theta_2 \\ \end{pmatrix} \\[1ex] = & \begin{pmatrix} \cos \theta _1\cos \theta _2 - \sin \theta _1 \sin \theta _2 & \cos \theta _1\sin \theta _2+ \sin \theta _1\cos \theta _2 \\ -\sin \theta _1\cos \theta _2-\cos \theta _1\sin \theta _2 & -\sin \theta _1\sin \theta _2 + \cos \theta _1\cos \theta _2 \\ \end{pmatrix} \\[1ex] = & \begin{pmatrix} \cos \left( \theta _1 + \theta _2 \right) & \sin \left( \theta _1 + \theta _2 \right) \\ -\sin \left( \theta _1 + \theta _2 \right) & \cos \left( \theta _1 + \theta _2 \right) \\ \end{pmatrix}. \end{align*}

Therefore, \begin{align*} A^{2023} & = \begin{pmatrix} \cos (2023\theta) & \sin (2023\theta ) \\[1ex] -\sin (2023\theta ) & \cos (2023\theta) \\ \end{pmatrix} \\[1ex] & = \begin{pmatrix} \cos \left( 2023\times \frac{2\pi}{17} \right) & \sin \left( 2023\times \frac{2\pi}{17} \right) \\[1ex] -\sin \left( 2023\times \frac{2\pi}{17} \right) & \cos \left( 2023\times \frac{2\pi}{17} \right) \\ \end{pmatrix} \\[1ex] & = \begin{pmatrix} \cos \left( 119\times 2\pi \right) & \sin \left( 119\times 2\pi \right) \\[1ex] -\sin \left( 119\times 2\pi \right) & \cos \left( 119\times 2\pi \right) \\ \end{pmatrix}\\[1ex] & = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}. \end{align*}