You're on a roll — Step 1: 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) $c(4) + d(2) = 1 \implies 4c + 2d = 1$ Step 2: Solve the system of equations for $a$ and $b$. 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): $$ \frac{1}{2} + 2b = \frac{1}{2} $$ $$ 2b = 0 $$ $$ b = 0 $$ Step 3: Solve the system of equations for $c$ and $d$. 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): $$ 0 + 2d = 1 $$ $$ 2d = 1 $$ $$ d = \frac{1}{2} $$ Step 4: Form the transformation matrix M. 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 $\boxed{\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{pmatrix}}$. What's next?