1. **Problem:** Quadrilateral WXYZ is mapped onto W'X'Y'Z' by transformation matrix $$P=\begin{pmatrix}1 & 3 \\ -3 & 1\end{pmatrix}$$. Given coordinates of W'X'Y'Z' as $$W'(3,-2), X'(1,-10), Y'(4,-12), Z'(12,-15)$$, find coordinates of WXYZ. 2. **Formula:** The transformation is $$P \times \textbf{WXYZ} = \textbf{W'X'Y'Z'}$$. To find WXYZ, multiply both sides by $$P^{-1}$$: $$\textbf{WXYZ} = P^{-1} \times \textbf{W'X'Y'Z'}$$. 3. **Find $$P^{-1}$$:** $$\det(P) = (1)(1) - (3)(-3) = 1 + 9 = 10$$ $$P^{-1} = \frac{1}{10} \begin{pmatrix}1 & -3 \\ 3 & 1\end{pmatrix}$$ 4. **Coordinates matrix of W'X'Y'Z':** $$\begin{pmatrix}3 & 1 & 4 & 12 \\ -2 & -10 & -12 & -15\end{pmatrix}$$ 5. **Calculate WXYZ:** $$\textbf{WXYZ} = \frac{1}{10} \begin{pmatrix}1 & -3 \\ 3 & 1\end{pmatrix} \times \begin{pmatrix}3 & 1 & 4 & 12 \\ -2 & -10 & -12 & -15\end{pmatrix}$$ 6. Multiply matrices: First row: $$1\times3 + (-3)\times(-2) = 3 + 6 = 9$$ $$1\times1 + (-3)\times(-10) = 1 + 30 = 31$$ $$1\times4 + (-3)\times(-12) = 4 + 36 = 40$$ $$1\times12 + (-3)\times(-15) = 12 + 45 = 57$$ Second row: $$3\times3 + 1\times(-2) = 9 - 2 = 7$$ $$3\times1 + 1\times(-10) = 3 - 10 = -7$$ $$3\times4 + 1\times(-12) = 12 - 12 = 0$$ $$3\times12 + 1\times(-15) = 36 - 15 = 21$$ 7. Divide each by 10: $$W = (\frac{9}{10}, \frac{7}{10}) = (0.9, 0.7)$$ $$X = (\frac{31}{10}, \frac{-7}{10}) = (3.1, -0.7)$$ $$Y = (\frac{40}{10}, 0) = (4, 0)$$ $$Z = (\frac{57}{10}, \frac{21}{10}) = (5.7, 2.1)$$ **Final coordinates:** $$W = (0.9, 0.7), X = (3.1, -0.7), Y = (4, 0), Z = (5.7, 2.1)$$