1. **Problem:** Find the inverse of the matrix $$A = \begin{pmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 3 & 4 & 2\end{pmatrix}$$ and solve the system: $$\begin{cases} x + 2y + z = 2 \\ 2x + 3y + z = 3 \\ 3x + 4y + 2z = 4 \end{cases}$$ 2. **Formula for inverse of a 3x3 matrix:** $$A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$$ where $\det(A)$ is the determinant and $\mathrm{adj}(A)$ is the adjugate matrix. 3. **Calculate determinant:** $$\det(A) = 1(3 \times 2 - 1 \times 4) - 2(2 \times 2 - 1 \times 3) + 1(2 \times 4 - 3 \times 3)$$ $$= 1(6 - 4) - 2(4 - 3) + 1(8 - 9) = 1 \times 2 - 2 \times 1 + 1 \times (-1) = 2 - 2 - 1 = -1$$ 4. **Find cofactors:** $$C_{11} = +(3 \times 2 - 1 \times 4) = 2$$ $$C_{12} = -(2 \times 2 - 1 \times 3) = -(4 - 3) = -1$$ $$C_{13} = +(2 \times 4 - 3 \times 3) = 8 - 9 = -1$$ $$C_{21} = -(2 \times 2 - 1 \times 3) = -1$$ $$C_{22} = +(1 \times 2 - 1 \times 3) = 2 - 3 = -1$$ $$C_{23} = -(1 \times 4 - 2 \times 3) = -(4 - 6) = 2$$ $$C_{31} = +(2 \times 1 - 3 \times 1) = 2 - 3 = -1$$ $$C_{32} = -(1 \times 1 - 2 \times 1) = -(1 - 2) = 1$$ $$C_{33} = +(1 \times 3 - 2 \times 2) = 3 - 4 = -1$$ 5. **Adjugate matrix (transpose of cofactor matrix):** $$\mathrm{adj}(A) = \begin{pmatrix} 2 & -1 & -1 \\ -1 & -1 & 2 \\ -1 & 1 & -1 \end{pmatrix}^T = \begin{pmatrix} 2 & -1 & -1 \\ -1 & -1 & 1 \\ -1 & 2 & -1 \end{pmatrix}$$ 6. **Inverse matrix:** $$A^{-1} = \frac{1}{-1} \begin{pmatrix} 2 & -1 & -1 \\ -1 & -1 & 1 \\ -1 & 2 & -1 \end{pmatrix} = \begin{pmatrix} -2 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -2 & 1 \end{pmatrix}$$ 7. **Solve system using $x = A^{-1}b$ where $b=\begin{pmatrix}2 \\ 3 \\ 4\end{pmatrix}$:** $$x = \begin{pmatrix} -2 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -2 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 \times 2 + 1 \times 3 + 1 \times 4 \\ 1 \times 2 + 1 \times 3 - 1 \times 4 \\ 1 \times 2 - 2 \times 3 + 1 \times 4 \end{pmatrix} = \begin{pmatrix} -4 + 3 + 4 \\ 2 + 3 - 4 \\ 2 - 6 + 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}$$ **Final answer:** $$A^{-1} = \begin{pmatrix} -2 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -2 & 1 \end{pmatrix}$$ $$\text{Solution: } x=3, y=1, z=0$$