1. **Stating the problem:** We have a transformation represented by the matrix $$\begin{pmatrix}1 & -1 \\ 0 & 2\end{pmatrix}$$. (a) Given the image of a point $A'$ is $(-2,4)$, find the coordinates of the original point $A$. (b) Find the area of the image of a rectangle with vertices $P(1,1)$, $Q(1,5)$, $R(4,5)$, and $S(4,1)$ under this transformation. 2. **Formula and rules:** The transformation matrix $M = \begin{pmatrix}1 & -1 \\ 0 & 2\end{pmatrix}$ acts on a point $\mathbf{x} = \begin{pmatrix}x \\ y\end{pmatrix}$ to produce $\mathbf{x'} = M \mathbf{x}$. To find $A$ from $A'$, we use the inverse matrix $M^{-1}$ such that $A = M^{-1} A'$. The area of the image of a shape under a linear transformation is the original area multiplied by the absolute value of the determinant of the transformation matrix. 3. **Find $A$ given $A' = (-2,4)$:** Calculate the determinant of $M$: $$\det(M) = (1)(2) - (0)(-1) = 2$$ Find the inverse matrix $M^{-1}$: $$M^{-1} = \frac{1}{\det(M)} \begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix} = \frac{1}{2} \begin{pmatrix}2 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}1 & \frac{1}{2} \\ 0 & \frac{1}{2}\end{pmatrix}$$ Apply $M^{-1}$ to $A' = \begin{pmatrix}-2 \\ 4\end{pmatrix}$: $$A = M^{-1} A' = \begin{pmatrix}1 & \frac{1}{2} \\ 0 & \frac{1}{2}\end{pmatrix} \begin{pmatrix}-2 \\ 4\end{pmatrix} = \begin{pmatrix}1 \times (-2) + \frac{1}{2} \times 4 \\ 0 \times (-2) + \frac{1}{2} \times 4\end{pmatrix} = \begin{pmatrix}-2 + 2 \\ 0 + 2\end{pmatrix} = \begin{pmatrix}0 \\ 2\end{pmatrix}$$ So, the coordinates of $A$ are $(0,2)$. 4. **Find the area of the image of the rectangle:** Calculate the original area of rectangle $PQRS$: Length $= 4 - 1 = 3$ Width $= 5 - 1 = 4$ Original area $= 3 \times 4 = 12$ The area of the image is: $$\text{Area}_{image} = |\det(M)| \times \text{Area}_{original} = 2 \times 12 = 24$$ **Final answers:** (a) Coordinates of $A$ are $(0,2)$. (b) Area of the image of the rectangle is $24$.