1. **Problem statement:** Solve the system of linear equations $$\begin{cases} x_1 + x_2 + 2x_3 = -1 \\ 2x_1 - x_2 + 2x_3 = -4 \\ 4x_1 + x_2 + 4x_3 = -2 \end{cases}$$ using a) Gaussian elimination and b) Cramer's rule. --- ### a) Gaussian elimination method 2. Write the augmented matrix: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 2 & -1 & 2 & -4 \\ 4 & 1 & 4 & -2 \end{array}\right]$$ 3. Use row operations to get upper triangular form. - Replace $R_2$ by $R_2 - 2R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ \cancel{2} - 2\times 1 & -1 - 2\times 1 & 2 - 2\times 2 & -4 - 2\times (-1) \\ 4 & 1 & 4 & -2 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 0 & -3 & -2 & -2 \\ 4 & 1 & 4 & -2 \end{array}\right]$$ - Replace $R_3$ by $R_3 - 4R_1$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 0 & -3 & -2 & -2 \\ \cancel{4} - 4\times 1 & 1 - 4\times 1 & 4 - 4\times 2 & -2 - 4\times (-1) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 0 & -3 & -2 & -2 \\ 0 & -3 & -4 & 2 \end{array}\right]$$ - Replace $R_3$ by $R_3 - R_2$: $$\left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 0 & -3 & -2 & -2 \\ 0 & \cancel{-3} - (-3) & -4 - (-2) & 2 - (-2) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 1 & 2 & -1 \\ 0 & -3 & -2 & -2 \\ 0 & 0 & -2 & 4 \end{array}\right]$$ 4. Back substitution: - From $R_3$: $-2x_3 = 4 \Rightarrow x_3 = \frac{4}{-2} = -2$ - From $R_2$: $-3x_2 - 2x_3 = -2 \Rightarrow -3x_2 - 2(-2) = -2 \Rightarrow -3x_2 + 4 = -2 \Rightarrow -3x_2 = -6 \Rightarrow x_2 = 2$ - From $R_1$: $x_1 + x_2 + 2x_3 = -1 \Rightarrow x_1 + 2 + 2(-2) = -1 \Rightarrow x_1 + 2 - 4 = -1 \Rightarrow x_1 - 2 = -1 \Rightarrow x_1 = 1$ --- ### b) Cramer's rule 5. Define coefficient matrix $A$ and constants vector $\mathbf{b}$: $$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & -1 & 2 \\ 4 & 1 & 4 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -1 \\ -4 \\ -2 \end{bmatrix}$$ 6. Calculate determinant $\det(A)$: $$\det(A) = 1 \times \begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} = (-1)(4) - (2)(1) = -4 - 2 = -6$$ $$\begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} = 2 \times 4 - 2 \times 4 = 8 - 8 = 0$$ $$\begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2 \times 1 - (-1) \times 4 = 2 + 4 = 6$$ So, $$\det(A) = 1 \times (-6) - 1 \times 0 + 2 \times 6 = -6 + 0 + 12 = 6$$ 7. Calculate determinants for $x_1, x_2, x_3$ replacing columns of $A$ with $\mathbf{b}$: - $\det(A_{x_1})$ (replace first column): $$\begin{vmatrix} -1 & 1 & 2 \\ -4 & -1 & 2 \\ -2 & 1 & 4 \end{vmatrix} = -1 \times \begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} - 1 \times \begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} -4 & -1 \\ -2 & 1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -1 & 2 \\ 1 & 4 \end{vmatrix} = -6$$ $$\begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} = (-4)(4) - (2)(-2) = -16 + 4 = -12$$ $$\begin{vmatrix} -4 & -1 \\ -2 & 1 \end{vmatrix} = (-4)(1) - (-1)(-2) = -4 - 2 = -6$$ So, $$\det(A_{x_1}) = -1 \times (-6) - 1 \times (-12) + 2 \times (-6) = 6 + 12 - 12 = 6$$ - $\det(A_{x_2})$ (replace second column): $$\begin{vmatrix} 1 & -1 & 2 \\ 2 & -4 & 2 \\ 4 & -2 & 4 \end{vmatrix} = 1 \times \begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} - (-1) \times \begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} + 2 \times \begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -4 & 2 \\ -2 & 4 \end{vmatrix} = -12$$ $$\begin{vmatrix} 2 & 2 \\ 4 & 4 \end{vmatrix} = 0$$ $$\begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix} = 2 \times (-2) - (-4) \times 4 = -4 + 16 = 12$$ So, $$\det(A_{x_2}) = 1 \times (-12) - (-1) \times 0 + 2 \times 12 = -12 + 0 + 24 = 12$$ - $\det(A_{x_3})$ (replace third column): $$\begin{vmatrix} 1 & 1 & -1 \\ 2 & -1 & -4 \\ 4 & 1 & -2 \end{vmatrix} = 1 \times \begin{vmatrix} -1 & -4 \\ 1 & -2 \end{vmatrix} - 1 \times \begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix} + (-1) \times \begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix}$$ Calculate minors: $$\begin{vmatrix} -1 & -4 \\ 1 & -2 \end{vmatrix} = (-1)(-2) - (-4)(1) = 2 + 4 = 6$$ $$\begin{vmatrix} 2 & -4 \\ 4 & -2 \end{vmatrix} = 2 \times (-2) - (-4) \times 4 = -4 + 16 = 12$$ $$\begin{vmatrix} 2 & -1 \\ 4 & 1 \end{vmatrix} = 2 \times 1 - (-1) \times 4 = 2 + 4 = 6$$ So, $$\det(A_{x_3}) = 1 \times 6 - 1 \times 12 + (-1) \times 6 = 6 - 12 - 6 = -12$$ 8. Calculate variables: $$x_1 = \frac{\det(A_{x_1})}{\det(A)} = \frac{6}{6} = 1$$ $$x_2 = \frac{\det(A_{x_2})}{\det(A)} = \frac{12}{6} = 2$$ $$x_3 = \frac{\det(A_{x_3})}{\det(A)} = \frac{-12}{6} = -2$$ --- **Final answer:** $$\boxed{(x_1, x_2, x_3) = (1, 2, -2)}$$