Set up the matrix multiplication for points A and B.

Let the transformation matrix be $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$.
The transformation maps an original point $(x, y)$ to an image point $(x', y')$ such that:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x' \\ y' \end{pmatrix}$$

For point A $(1,2)$ mapping to A' $(\frac{1}{2}, 1)$:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ 1 \end{pmatrix}$$

This gives the equations:

1. $a(1) + b(2) = \frac{1}{2} \implies a + 2b = \frac{1}{2}$
2. $c(1) + d(2) = 1 \implies c + 2d = 1$

For point B $(4,2)$ mapping to B' $(2,1)$:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

This gives the equations: 3) $a(4) + b(2) = 2 \implies 4a + 2b = 2$ 4) $c(4) + d(2) = 1 \implies 4c + 2d = 1$

We use equations (1) and (3):

$$a + 2b = \frac{1}{2} \quad (1)$$ $$4a + 2b = 2 \quad (3)$$

Subtract equation (1) from equation (3):

$$(4a + 2b) - (a + 2b) = 2 - \frac{1}{2}$$ $$3a = \frac{4}{2} - \frac{1}{2}$$ $$3a = \frac{3}{2}$$ $$a = \frac{3}{2} \times \frac{1}{3}$$ $$a = \frac{1}{2}$$

Substitute $a = \frac{1}{2}$ into equation (1):

$$\frac{1}{2} + 2b = \frac{1}{2}$$ $$2b = 0$$ $$b = 0$$

We use equations (2) and (4):

$$c + 2d = 1 \quad (2)$$ $$4c + 2d = 1 \quad (4)$$

Subtract equation (2) from equation (4):

$$(4c + 2d) - (c + 2d) = 1 - 1$$ $$3c = 0$$ $$c = 0$$

Substitute $c = 0$ into equation (2):

$$0 + 2d = 1$$ $$2d = 1$$ $$d = \frac{1}{2}$$

Using the values $a = \frac{1}{2}$, $b = 0$, $c = 0$, $d = \frac{1}{2}$:

$$M = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}$$

The transformation matrix is $\boxed{\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}}$.