1. **Stating the problem:** We have a triangle with vertices $P(-3,2)$, $Q(0,-1)$, and $R(2,-1)$. A transformation matrix $M$ maps these points to $P^\prime(-7,2)$, $Q^\prime(2,-1)$, and $R^\prime(4,-1)$ respectively. We need to find the inverse matrix $M^{-1}$ that maps $P^\prime, Q^\prime, R^\prime$ back to $P, Q, R$. 2. **Formula and approach:** The transformation can be written as $$\begin{bmatrix}P^\prime & Q^\prime & R^\prime\end{bmatrix} = M \begin{bmatrix}P & Q & R\end{bmatrix}$$ where each matrix contains the coordinates as columns: $$\begin{bmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{bmatrix}$$ To find $M$, we use: $$M = \begin{bmatrix}P^\prime & Q^\prime & R^\prime\end{bmatrix} \cdot \begin{bmatrix}P & Q & R\end{bmatrix}^{-1}$$ Then, $$M^{-1} = \left(M\right)^{-1} = \begin{bmatrix}P & Q & R\end{bmatrix} \cdot \begin{bmatrix}P^\prime & Q^\prime & R^\prime\end{bmatrix}^{-1}$$ 3. **Construct matrices:** $$A = \begin{bmatrix}-3 & 0 & 2 \\ 2 & -1 & -1\end{bmatrix}, \quad B = \begin{bmatrix}-7 & 2 & 4 \\ 2 & -1 & -1\end{bmatrix}$$ 4. **Calculate $A^{-1}$:** First, find determinant of $A$: $$\det(A) = (-3)(-1) - (0)(2) = 3$$ Since $A$ is $2 \times 3$, not square, we consider only the $2 \times 2$ part for transformation. Actually, for affine transformations, we consider $2 \times 2$ matrices for linear parts. Here, points are 2D vectors, so we consider $2 \times 2$ matrices formed by vectors $P, Q$ as columns and similarly for $P^\prime, Q^\prime$. Use points $P$ and $Q$ only for matrix $A$ and $P^\prime$ and $Q^\prime$ for matrix $B$: $$A = \begin{bmatrix}-3 & 0 \\ 2 & -1\end{bmatrix}, \quad B = \begin{bmatrix}-7 & 2 \\ 2 & -1\end{bmatrix}$$ Calculate $A^{-1}$: $$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix}-1 & 0 \\ -2 & -3\end{bmatrix}$$ But this is incorrect; the formula for inverse of $2 \times 2$ matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is: $$\frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$ So, $$\det(A) = (-3)(-1) - (0)(2) = 3$$ $$A^{-1} = \frac{1}{3} \begin{bmatrix}-1 & 0 \\ -2 & -3\end{bmatrix}$$ 5. **Calculate $M$:** $$M = B \cdot A^{-1} = \begin{bmatrix}-7 & 2 \\ 2 & -1\end{bmatrix} \cdot \frac{1}{3} \begin{bmatrix}-1 & 0 \\ -2 & -3\end{bmatrix} = \frac{1}{3} \begin{bmatrix}(-7)(-1) + 2(-2) & (-7)(0) + 2(-3) \\ 2(-1) + (-1)(-2) & 2(0) + (-1)(-3)\end{bmatrix}$$ Calculate each element: $$= \frac{1}{3} \begin{bmatrix}7 - 4 & 0 - 6 \\ -2 + 2 & 0 + 3\end{bmatrix} = \frac{1}{3} \begin{bmatrix}3 & -6 \\ 0 & 3\end{bmatrix} = \begin{bmatrix}1 & -2 \\ 0 & 1\end{bmatrix}$$ 6. **Find $M^{-1}$:** For $M = \begin{bmatrix}1 & -2 \\ 0 & 1\end{bmatrix}$, $$\det(M) = 1 \times 1 - (-2) \times 0 = 1$$ $$M^{-1} = \frac{1}{1} \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$$ **Final answer:** $$M^{-1} = \begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}$$