1. **Problem:** Solve the system of equations using matrix method: $$\begin{cases} x + 2y = 7 \\ x + 3y + 4z = 16 \\ x + y + z = 7 \end{cases}$$ 2. **Step 1: Write the system in matrix form $AX = B$ where** $$A = \begin{bmatrix} 1 & 2 & 0 \\ 1 & 3 & 4 \\ 1 & 1 & 1 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 7 \\ 16 \\ 7 \end{bmatrix}$$ 3. **Step 2: Find the determinant of matrix $A$ to check if inverse exists:** $$\det(A) = 1 \times \begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} - 2 \times \begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} + 0 \times \begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix}$$ Calculate minors: $$= 1 \times (3 \times 1 - 4 \times 1) - 2 \times (1 \times 1 - 4 \times 1) + 0$$ $$= 1 \times (3 - 4) - 2 \times (1 - 4)$$ $$= 1 \times (-1) - 2 \times (-3) = -1 + 6 = 5$$ Since $\det(A) = 5 \neq 0$, inverse exists. 4. **Step 3: Find the inverse of $A$, $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$** Calculate cofactors matrix: $$C = \begin{bmatrix} +\begin{vmatrix} 3 & 4 \\ 1 & 1 \end{vmatrix} & -\begin{vmatrix} 1 & 4 \\ 1 & 1 \end{vmatrix} & +\begin{vmatrix} 1 & 3 \\ 1 & 1 \end{vmatrix} \\ -\begin{vmatrix} 2 & 0 \\ 1 & 1 \end{vmatrix} & +\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \\ 1 & 1 \end{vmatrix} \\ +\begin{vmatrix} 2 & 0 \\ 3 & 4 \end{vmatrix} & -\begin{vmatrix} 1 & 0 \\ 1 & 4 \end{vmatrix} & +\begin{vmatrix} 1 & 2 \\ 1 & 3 \end{vmatrix} \end{bmatrix}$$ Calculate each minor: $$C = \begin{bmatrix} (3 \times 1 - 4 \times 1) & -(1 \times 1 - 4 \times 1) & (1 \times 1 - 3 \times 1) \\ -(2 \times 1 - 0 \times 1) & (1 \times 1 - 0 \times 1) & -(1 \times 1 - 2 \times 1) \\ (2 \times 4 - 0 \times 3) & -(1 \times 4 - 0 \times 1) & (1 \times 3 - 2 \times 1) \end{bmatrix}$$ Simplify: $$C = \begin{bmatrix} (3 - 4) & -(1 - 4) & (1 - 3) \\ -(2 - 0) & (1 - 0) & -(1 - 2) \\ (8 - 0) & -(4 - 0) & (3 - 2) \end{bmatrix} = \begin{bmatrix} -1 & 3 & -2 \\ -2 & 1 & 1 \\ 8 & -4 & 1 \end{bmatrix}$$ 5. **Step 4: Transpose the cofactor matrix to get adjugate:** $$\text{adj}(A) = C^T = \begin{bmatrix} -1 & -2 & 8 \\ 3 & 1 & -4 \\ -2 & 1 & 1 \end{bmatrix}$$ 6. **Step 5: Calculate $A^{-1}$:** $$A^{-1} = \frac{1}{5} \begin{bmatrix} -1 & -2 & 8 \\ 3 & 1 & -4 \\ -2 & 1 & 1 \end{bmatrix}$$ 7. **Step 6: Find $X = A^{-1} B$:** $$X = \frac{1}{5} \begin{bmatrix} -1 & -2 & 8 \\ 3 & 1 & -4 \\ -2 & 1 & 1 \end{bmatrix} \begin{bmatrix} 7 \\ 16 \\ 7 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} (-1)(7) + (-2)(16) + 8(7) \\ 3(7) + 1(16) + (-4)(7) \\ -2(7) + 1(16) + 1(7) \end{bmatrix}$$ Calculate each element: $$= \frac{1}{5} \begin{bmatrix} -7 - 32 + 56 \\ 21 + 16 - 28 \\ -14 + 16 + 7 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 17 \\ 9 \\ 9 \end{bmatrix}$$ 8. **Step 7: Simplify to get values of $x, y, z$:** $$x = \frac{17}{5} = 3.4, \quad y = \frac{9}{5} = 1.8, \quad z = \frac{9}{5} = 1.8$$ **Final answer:** $$\boxed{x = 3.4, \quad y = 1.8, \quad z = 1.8}$$