1. **State the problem:** We are given a linear system: $$\begin{cases} x + y = 1 \\ -x + 2y = 2 \end{cases}$$ with matrix $$A = \begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}$$ and its inverse $$A^{-1} = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ We want to find $x$ and $y$ using $A^{-1}$. 2. **Formula used:** The solution vector $\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix}$ is given by $$\mathbf{x} = A^{-1} \mathbf{b}$$ where $$\mathbf{b} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$ from the right side of the system. 3. **Find $A^{-1}$:** The inverse of a 2x2 matrix $$A = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$$ is $$A^{-1} = \frac{1}{\det(A)} \begin{bmatrix} s & -q \\ -r & p \end{bmatrix}$$ where $$\det(A) = ps - rq$$ For our matrix, $$p=1, q=1, r=-1, s=2$$ Calculate determinant: $$\det(A) = (1)(2) - (-1)(1) = 2 + 1 = 3$$ So, $$A^{-1} = \frac{1}{3} \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{bmatrix}$$ Thus, $$a = \frac{2}{3}, b = -\frac{1}{3}, c = \frac{1}{3}, d = \frac{1}{3}$$ 4. **Solve for $x$ and $y$:** $$\mathbf{x} = A^{-1} \mathbf{b} = \begin{bmatrix} \frac{2}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$ Calculate each component: $$x = \frac{2}{3} \times 1 + \left(-\frac{1}{3}\right) \times 2 = \frac{2}{3} - \frac{2}{3} = 0$$ $$y = \frac{1}{3} \times 1 + \frac{1}{3} \times 2 = \frac{1}{3} + \frac{2}{3} = 1$$ 5. **Final answer:** $$x = 0, \quad y = 1$$ This satisfies the original system.