1. **Problem Statement:** Given the system of linear equations: $$\begin{cases} 3x + 2y + z = 2 \\ x + 3y + 2z = 1 \\ -4y - 3z = -2 \end{cases}$$ 2. **Part (a): Write the system in matrix form $A \mathbf{X} = \mathbf{B}$** The coefficient matrix $A$, variable vector $\mathbf{X}$, and constant vector $\mathbf{B}$ are: $$A = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 3 & 2 \\ 0 & -4 & -3 \end{bmatrix}, \quad \mathbf{X} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{B} = \begin{bmatrix} 2 \\ 1 \\ -2 \end{bmatrix}$$ So the system is: $$A \mathbf{X} = \mathbf{B}$$ 3. **Part (b): Show that the system has a non-trivial solution** A non-trivial solution exists if the determinant of $A$ is zero (system is singular) or if the system is consistent with free variables. Calculate $\det(A)$: $$\det(A) = 3 \begin{vmatrix} 3 & 2 \\ -4 & -3 \end{vmatrix} - 2 \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} + 1 \begin{vmatrix} 1 & 3 \\ 0 & -4 \end{vmatrix}$$ Calculate minors: $$= 3(3 \times -3 - 2 \times -4) - 2(1 \times -3 - 0) + 1(1 \times -4 - 0)$$ $$= 3(-9 + 8) - 2(-3) + 1(-4) = 3(-1) + 6 - 4 = -3 + 6 - 4 = -1$$ Since $\det(A) = -1 \neq 0$, the system has only the trivial solution for the homogeneous case. 4. **Part (c): Solve the system by reducing the augmented matrix $(A|B)$ to echelon form** Write the augmented matrix: $$\left[ \begin{array}{ccc|c} 3 & 2 & 1 & 2 \\ 1 & 3 & 2 & 1 \\ 0 & -4 & -3 & -2 \end{array} \right]$$ Step 1: Make leading 1 in row 1 by dividing row 1 by 3: $$R1 \to \frac{1}{3} R1 = \left[1, \frac{2}{3}, \frac{1}{3}, \frac{2}{3}\right]$$ Step 2: Eliminate $x$ in row 2: $$R2 \to R2 - R2[1] \times R1 = R2 - 1 \times R1$$ $$= [1, 3, 2, 1] - [1, \frac{2}{3}, \frac{1}{3}, \frac{2}{3}] = [0, \frac{7}{3}, \frac{5}{3}, \frac{1}{3}]$$ Step 3: Row 3 already has 0 in $x$ position. Augmented matrix now: $$\left[ \begin{array}{ccc|c} 1 & \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{7}{3} & \frac{5}{3} & \frac{1}{3} \\ 0 & -4 & -3 & -2 \end{array} \right]$$ Step 4: Make leading 1 in row 2 by multiplying by $\frac{3}{7}$: $$R2 \to \frac{3}{7} R2 = \left[0, 1, \frac{5}{7}, \frac{1}{7}\right]$$ Step 5: Eliminate $y$ in row 3: $$R3 \to R3 + 4 \times R2 = [0, -4, -3, -2] + 4 \times [0, 1, \frac{5}{7}, \frac{1}{7}] = [0, 0, -3 + \frac{20}{7}, -2 + \frac{4}{7}] = [0, 0, -\frac{1}{7}, -\frac{10}{7}]$$ Step 6: Make leading 1 in row 3 by multiplying by $-7$: $$R3 \to -7 R3 = [0, 0, 1, 10]$$ Step 7: Back substitution to eliminate $z$ from rows 1 and 2: Row 2: $$R2 \to R2 - \frac{5}{7} R3 = [0, 1, \frac{5}{7}, \frac{1}{7}] - \frac{5}{7} [0, 0, 1, 10] = [0, 1, 0, \frac{1}{7} - \frac{50}{7}] = [0, 1, 0, -7]$$ Row 1: $$R1 \to R1 - \frac{1}{3} R3 = [1, \frac{2}{3}, \frac{1}{3}, \frac{2}{3}] - \frac{1}{3} [0, 0, 1, 10] = [1, \frac{2}{3}, 0, \frac{2}{3} - \frac{10}{3}] = [1, \frac{2}{3}, 0, -\frac{8}{3}]$$ Step 8: Eliminate $y$ from row 1: $$R1 \to R1 - \frac{2}{3} R2 = [1, \frac{2}{3}, 0, -\frac{8}{3}] - \frac{2}{3} [0, 1, 0, -7] = [1, 0, 0, -\frac{8}{3} + \frac{14}{3}] = [1, 0, 0, 2]$$ **Final solution:** $$x = 2, \quad y = -7, \quad z = 10$$ **Summary:** - Matrix form: $A \mathbf{X} = \mathbf{B}$ - Determinant $\neq 0$, so unique solution exists - Solution by row reduction: $x=2$, $y=-7$, $z=10$