1. **State the problem:** Solve the system of equations using matrices: $$\begin{cases} x + 3y + z = 3 \\ x + 5y + 5z = 1 \\ 2x + 6y + 3z = 8 \end{cases}$$ 2. **Write the system in matrix form:** $$A = \begin{bmatrix} 1 & 3 & 1 \\ 1 & 5 & 5 \\ 2 & 6 & 3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 3 \\ 1 \\ 8 \end{bmatrix}$$ The system is $A\mathbf{x} = \mathbf{b}$. 3. **Find the inverse of matrix $A$ if it exists:** Calculate the determinant of $A$: $$\det(A) = 1(5 \cdot 3 - 5 \cdot 6) - 3(1 \cdot 3 - 5 \cdot 2) + 1(1 \cdot 6 - 5 \cdot 2)$$ $$= 1(15 - 30) - 3(3 - 10) + 1(6 - 10) = 1(-15) - 3(-7) + 1(-4) = -15 + 21 - 4 = 2$$ Since $\det(A) = 2 \neq 0$, $A$ is invertible. 4. **Calculate $A^{-1}$ (inverse of $A$):** Using the formula $A^{-1} = \frac{1}{\det(A)} \mathrm{adj}(A)$, where $\mathrm{adj}(A)$ is the adjugate matrix. Calculate cofactors: $$C_{11} = \det \begin{bmatrix} 5 & 5 \\ 6 & 3 \end{bmatrix} = 5 \cdot 3 - 5 \cdot 6 = 15 - 30 = -15$$ $$C_{12} = -\det \begin{bmatrix} 1 & 5 \\ 2 & 3 \end{bmatrix} = -(1 \cdot 3 - 5 \cdot 2) = -(3 - 10) = 7$$ $$C_{13} = \det \begin{bmatrix} 1 & 5 \\ 2 & 6 \end{bmatrix} = 1 \cdot 6 - 5 \cdot 2 = 6 - 10 = -4$$ $$C_{21} = -\det \begin{bmatrix} 3 & 1 \\ 6 & 3 \end{bmatrix} = -(3 \cdot 3 - 1 \cdot 6) = -(9 - 6) = -3$$ $$C_{22} = \det \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} = 1 \cdot 3 - 1 \cdot 2 = 3 - 2 = 1$$ $$C_{23} = -\det \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix} = -(1 \cdot 6 - 3 \cdot 2) = -(6 - 6) = 0$$ $$C_{31} = \det \begin{bmatrix} 3 & 1 \\ 5 & 5 \end{bmatrix} = 3 \cdot 5 - 1 \cdot 5 = 15 - 5 = 10$$ $$C_{32} = -\det \begin{bmatrix} 1 & 1 \\ 1 & 5 \end{bmatrix} = -(1 \cdot 5 - 1 \cdot 1) = -(5 - 1) = -4$$ $$C_{33} = \det \begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix} = 1 \cdot 5 - 3 \cdot 1 = 5 - 3 = 2$$ 5. **Form the cofactor matrix and transpose it to get adjugate:** $$\mathrm{adj}(A) = \begin{bmatrix} C_{11} & C_{21} & C_{31} \\ C_{12} & C_{22} & C_{32} \\ C_{13} & C_{23} & C_{33} \end{bmatrix} = \begin{bmatrix} -15 & -3 & 10 \\ 7 & 1 & -4 \\ -4 & 0 & 2 \end{bmatrix}$$ 6. **Calculate $A^{-1}$:** $$A^{-1} = \frac{1}{2} \begin{bmatrix} -15 & -3 & 10 \\ 7 & 1 & -4 \\ -4 & 0 & 2 \end{bmatrix} = \begin{bmatrix} -\frac{15}{2} & -\frac{3}{2} & 5 \\ \frac{7}{2} & \frac{1}{2} & -2 \\ -2 & 0 & 1 \end{bmatrix}$$ 7. **Find solution vector $\mathbf{x} = A^{-1} \mathbf{b}$:** $$\mathbf{x} = \begin{bmatrix} -\frac{15}{2} & -\frac{3}{2} & 5 \\ \frac{7}{2} & \frac{1}{2} & -2 \\ -2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \\ 8 \end{bmatrix}$$ Calculate each component: $$x = -\frac{15}{2} \cdot 3 - \frac{3}{2} \cdot 1 + 5 \cdot 8 = -\frac{45}{2} - \frac{3}{2} + 40 = -24 + 40 = 16$$ $$y = \frac{7}{2} \cdot 3 + \frac{1}{2} \cdot 1 - 2 \cdot 8 = \frac{21}{2} + \frac{1}{2} - 16 = 11 - 16 = -5$$ $$z = -2 \cdot 3 + 0 \cdot 1 + 1 \cdot 8 = -6 + 0 + 8 = 2$$ 8. **Final answer:** $$\boxed{x = 16, \quad y = -5, \quad z = 2}$$