If $\text{A}=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, then ${\text{A}}^n$ is equal to:

Solution

1. The given matrix is $\text{A}=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$, which represents a rotation transformation by an angle $\theta$.
2. By the property of rotation matrices, applying the transformation $n$ times corresponds to a rotation by the angle $n\theta$.
3. Therefore, ${\text{A}}^n=\left(\begin{array}{cc}\cos (n\theta )& -\sin (n\theta )\\ \sin (n\theta )& \cos (n\theta )\end{array}\right)$.
4. Hence the answer is (A).