State the general rotation matrix for an anticlockwise rotation.

The matrix for an anticlockwise rotation of a point $(x, y)$ about the origin through an angle $\theta$ is given by:

$$R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$

The transformed coordinates $(x', y')$ are found by:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

This matrix is shown at the bottom of the provided image.

To verify the specific case of a $270^\circ$ anticlockwise rotation shown in the image, we substitute $\theta = 270^\circ$ into the general rotation matrix:

$$R(270^\circ) = \begin{pmatrix} \cos(270^\circ) & -\sin(270^\circ) \\ \sin(270^\circ) & \cos(270^\circ) \end{pmatrix}$$

We know that $\cos(270^\circ) = 0$ and $\sin(270^\circ) = -1$.
Substitute these values:

$$R(270^\circ) = \begin{pmatrix} 0 & -(-1) \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$

This matches the rotation matrix shown in the image for the $270^\circ$ rotation.

Now, apply this matrix to a point $\begin{pmatrix} x \\ y \end{pmatrix}$:

$$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (0)(x) + (1)(y) \\ (-1)(x) + (0)(y) \end{pmatrix} = \begin{pmatrix} y \\ -x \end{pmatrix}$$

This result $\begin{pmatrix} y \\ -x \end{pmatrix}$ also matches the transformed coordinates shown in the image for a $270^\circ$ anticlockwise rotation of $\begin{pmatrix} x \\ y \end{pmatrix}$.

The general rotation matrix for an anticlockwise rotation through an angle $\theta$ is $\boxed{\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}}$.
The specific rotation matrix for an anticlockwise rotation through $270^\circ$ is $\boxed{\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}}$.