1. **State the problem:** We are given the equation $$15 \cdot (7A)^{-1} = \begin{bmatrix}-3 & 7 \\ 1 & -2\end{bmatrix}$$ and need to find the matrix $$A$$. 2. **Recall the formula for the inverse of a matrix:** For a 2x2 matrix $$M = \begin{bmatrix}a & b \\ c & d\end{bmatrix}$$, its inverse is given by $$M^{-1} = \frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a\end{bmatrix}$$ provided that $$ad - bc \neq 0$$. 3. **Rewrite the given equation:** $$15 \cdot (7A)^{-1} = \begin{bmatrix}-3 & 7 \\ 1 & -2\end{bmatrix}$$ Divide both sides by 15: $$ (7A)^{-1} = \frac{1}{15} \begin{bmatrix}-3 & 7 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}-\frac{1}{5} & \frac{7}{15} \\ \frac{1}{15} & -\frac{2}{15}\end{bmatrix}$$ 4. **Find the inverse of both sides:** Taking inverse on both sides, $$7A = \left( (7A)^{-1} \right)^{-1} = \left( \begin{bmatrix}-\frac{1}{5} & \frac{7}{15} \\ \frac{1}{15} & -\frac{2}{15}\end{bmatrix} \right)^{-1}$$ 5. **Calculate the determinant of the matrix on the right:** $$det = \left(-\frac{1}{5}\right) \cdot \left(-\frac{2}{15}\right) - \left(\frac{7}{15}\right) \cdot \left(\frac{1}{15}\right) = \frac{2}{75} - \frac{7}{225} = \frac{6}{225} - \frac{7}{225} = -\frac{1}{225}$$ 6. **Calculate the inverse matrix:** $$\left( \begin{bmatrix}-\frac{1}{5} & \frac{7}{15} \\ \frac{1}{15} & -\frac{2}{15}\end{bmatrix} \right)^{-1} = \frac{1}{det} \begin{bmatrix} -\frac{2}{15} & -\frac{7}{15} \\ -\frac{1}{15} & -\frac{1}{5} \end{bmatrix} = -225 \cdot \begin{bmatrix} -\frac{2}{15} & -\frac{7}{15} \\ -\frac{1}{15} & -\frac{1}{5} \end{bmatrix}$$ 7. **Multiply scalar and matrix:** $$= -225 \cdot \begin{bmatrix} -\frac{2}{15} & -\frac{7}{15} \\ -\frac{1}{15} & -\frac{1}{5} \end{bmatrix} = \begin{bmatrix} 30 & 105 \\ 15 & 45 \end{bmatrix}$$ 8. **Recall that $$7A = \begin{bmatrix}30 & 105 \\ 15 & 45\end{bmatrix}$$, so divide both sides by 7 to find $$A$$:** $$A = \frac{1}{7} \begin{bmatrix}30 & 105 \\ 15 & 45\end{bmatrix} = \begin{bmatrix} \frac{30}{7} & 15 \\ \frac{15}{7} & \frac{45}{7} \end{bmatrix}$$ **Final answer:** $$A = \begin{bmatrix} \frac{30}{7} & 15 \\ \frac{15}{7} & \frac{45}{7} \end{bmatrix}$$