1. **State the problem:** We are given the inverse of matrix $A$, denoted as $A^{-1} = \begin{bmatrix} 2 & -1 \\ 3 & 5 \end{bmatrix}$, and we need to find the original matrix $A$. 2. **Recall the formula:** For a 2x2 matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, its inverse is given by: $$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ where $ad - bc \neq 0$ is the determinant of $A$. 3. **Set up equations:** Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Then: $$A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ 3 & 5 \end{bmatrix}$$ 4. **Equate elements:** From the matrices, we get: $$\frac{d}{ad - bc} = 2, \quad \frac{-b}{ad - bc} = -1, \quad \frac{-c}{ad - bc} = 3, \quad \frac{a}{ad - bc} = 5$$ 5. **Rewrite equations:** $$\frac{d}{\det(A)} = 2 \Rightarrow d = 2 \det(A)$$ $$\frac{-b}{\det(A)} = -1 \Rightarrow -b = -\det(A) \Rightarrow b = \det(A)$$ $$\frac{-c}{\det(A)} = 3 \Rightarrow -c = 3 \det(A) \Rightarrow c = -3 \det(A)$$ $$\frac{a}{\det(A)} = 5 \Rightarrow a = 5 \det(A)$$ 6. **Recall determinant:** $$\det(A) = ad - bc = a d - b c$$ Substitute $a, b, c, d$ in terms of $\det(A)$: $$\det(A) = (5 \det(A))(2 \det(A)) - (\det(A))(-3 \det(A)) = 10 \det(A)^2 + 3 \det(A)^2 = 13 \det(A)^2$$ 7. **Solve for $\det(A)$:** $$\det(A) = 13 \det(A)^2 \Rightarrow 13 \det(A)^2 - \det(A) = 0 \Rightarrow \det(A)(13 \det(A) - 1) = 0$$ Since $\det(A) \neq 0$, we have: $$13 \det(A) - 1 = 0 \Rightarrow \det(A) = \frac{1}{13}$$ 8. **Find matrix elements:** $$a = 5 \times \frac{1}{13} = \frac{5}{13}$$ $$b = \frac{1}{13}$$ $$c = -3 \times \frac{1}{13} = -\frac{3}{13}$$ $$d = 2 \times \frac{1}{13} = \frac{2}{13}$$ 9. **Final answer:** $$A = \begin{bmatrix} \frac{5}{13} & \frac{1}{13} \\ -\frac{3}{13} & \frac{2}{13} \end{bmatrix}$$ This is the original matrix $A$ whose inverse is given.