1. **Problem Statement:** Solve the system of linear equations: $$\begin{cases} 3x_1 + x_2 + x_3 = 8 \\ x_1 + 4x_2 + 2x_3 = 15 \\ 2x_1 + x_2 + 5x_3 = 19 \end{cases}$$ 2. **Method a: Cramer's Rule** - Formula: For system $AX = B$, solution $x_i = \frac{\det(A_i)}{\det(A)}$ where $A_i$ is matrix $A$ with $i$-th column replaced by $B$. - Calculate determinant of coefficient matrix $A$: $$A = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 4 & 2 \\ 2 & 1 & 5 \end{bmatrix}$$ $$\det(A) = 3(4 \times 5 - 2 \times 1) - 1(1 \times 5 - 2 \times 2) + 1(1 \times 1 - 4 \times 2)$$ $$= 3(20 - 2) - 1(5 - 4) + 1(1 - 8) = 3 \times 18 - 1 \times 1 + 1 \times (-7) = 54 - 1 - 7 = 46$$ - Replace columns and find determinants: $$A_1 = \begin{bmatrix} 8 & 1 & 1 \\ 15 & 4 & 2 \\ 19 & 1 & 5 \end{bmatrix}$$ $$\det(A_1) = 8(4 \times 5 - 2 \times 1) - 1(15 \times 5 - 2 \times 19) + 1(15 \times 1 - 4 \times 19)$$ $$= 8(20 - 2) - 1(75 - 38) + 1(15 - 76) = 8 \times 18 - 37 - 61 = 144 - 37 - 61 = 46$$ $$A_2 = \begin{bmatrix} 3 & 8 & 1 \\ 1 & 15 & 2 \\ 2 & 19 & 5 \end{bmatrix}$$ $$\det(A_2) = 3(15 \times 5 - 2 \times 19) - 8(1 \times 5 - 2 \times 2) + 1(1 \times 19 - 15 \times 2)$$ $$= 3(75 - 38) - 8(5 - 4) + 1(19 - 30) = 3 \times 37 - 8 \times 1 - 11 = 111 - 8 - 11 = 92$$ $$A_3 = \begin{bmatrix} 3 & 1 & 8 \\ 1 & 4 & 15 \\ 2 & 1 & 19 \end{bmatrix}$$ $$\det(A_3) = 3(4 \times 19 - 15 \times 1) - 1(1 \times 19 - 15 \times 2) + 8(1 \times 1 - 4 \times 2)$$ $$= 3(76 - 15) - 1(19 - 30) + 8(1 - 8) = 3 \times 61 + 11 - 56 = 183 + 11 - 56 = 138$$ - Solutions: $$x_1 = \frac{46}{46} = 1, \quad x_2 = \frac{92}{46} = 2, \quad x_3 = \frac{138}{46} = 3$$ 3. **Final answer:** $$\boxed{x_1 = 1, x_2 = 2, x_3 = 3}$$ The other methods (Gauss Elimination, Gauss-Jordan, Gauss-Seidel) will yield the same solution but are not shown here due to the guest rule limiting to the first question only.