1. **Problem statement:** Solve the system of linear equations: $$\begin{cases}-4x_2 + x_3 + 2x_4 = 8 \\ 2x_1 + 3x_2 + x_3 - x_4 = 3 \\ x_1 - 2x_2 + 4x_3 + 3x_4 = 19 \end{cases}$$ (i) Form the augmented matrix and reduce it to row-reduced echelon form (RREF). 2. **Form the augmented matrix:** $$\left[\begin{array}{cccc|c}0 & -4 & 1 & 2 & 8 \\ 2 & 3 & 1 & -1 & 3 \\ 1 & -2 & 4 & 3 & 19 \end{array}\right]$$ 3. **Row operations to get RREF:** - Swap $R_1$ and $R_3$ to get a leading 1 in the first row: $$R_1 \leftrightarrow R_3$$ $$\left[\begin{array}{cccc|c}1 & -2 & 4 & 3 & 19 \\ 2 & 3 & 1 & -1 & 3 \\ 0 & -4 & 1 & 2 & 8 \end{array}\right]$$ - Eliminate $x_1$ from $R_2$: $$R_2 \to R_2 - 2R_1$$ $$\left[\begin{array}{cccc|c}1 & -2 & 4 & 3 & 19 \\ 0 & 7 & -7 & -7 & -35 \\ 0 & -4 & 1 & 2 & 8 \end{array}\right]$$ - Eliminate $x_1$ from $R_3$ (already zero, no change). - Make leading coefficient in $R_2$ equal to 1: $$R_2 \to \frac{1}{7} R_2$$ $$\left[\begin{array}{cccc|c}1 & -2 & 4 & 3 & 19 \\ 0 & 1 & -1 & -1 & -5 \\ 0 & -4 & 1 & 2 & 8 \end{array}\right]$$ - Eliminate $x_2$ from $R_1$: $$R_1 \to R_1 + 2R_2$$ $$\left[\begin{array}{cccc|c}1 & 0 & 2 & 1 & 9 \\ 0 & 1 & -1 & -1 & -5 \\ 0 & -4 & 1 & 2 & 8 \end{array}\right]$$ - Eliminate $x_2$ from $R_3$: $$R_3 \to R_3 + 4R_2$$ $$\left[\begin{array}{cccc|c}1 & 0 & 2 & 1 & 9 \\ 0 & 1 & -1 & -1 & -5 \\ 0 & 0 & -3 & -2 & -12 \end{array}\right]$$ - Make leading coefficient in $R_3$ equal to 1: $$R_3 \to -\frac{1}{3} R_3$$ $$\left[\begin{array}{cccc|c}1 & 0 & 2 & 1 & 9 \\ 0 & 1 & -1 & -1 & -5 \\ 0 & 0 & 1 & \frac{2}{3} & 4 \end{array}\right]$$ - Eliminate $x_3$ from $R_1$: $$R_1 \to R_1 - 2R_3$$ $$\left[\begin{array}{cccc|c}1 & 0 & 0 & \frac{-1}{3} & 1 \\ 0 & 1 & -1 & -1 & -5 \\ 0 & 0 & 1 & \frac{2}{3} & 4 \end{array}\right]$$ - Eliminate $x_3$ from $R_2$: $$R_2 \to R_2 + R_3$$ $$\left[\begin{array}{cccc|c}1 & 0 & 0 & \frac{-1}{3} & 1 \\ 0 & 1 & 0 & -\frac{1}{3} & -1 \\ 0 & 0 & 1 & \frac{2}{3} & 4 \end{array}\right]$$ - Eliminate $x_4$ from $R_1$ and $R_2$ to get RREF: Let $R_1 \to R_1 + \frac{1}{3} R_2$: $$R_1 = \left[1, 0, 0, 0, 1 + \frac{1}{3}(-1)\right] = \left[1, 0, 0, 0, \frac{2}{3}\right]$$ $$\left[\begin{array}{cccc|c}1 & 0 & 0 & 0 & \frac{2}{3} \\ 0 & 1 & 0 & -\frac{1}{3} & -1 \\ 0 & 0 & 1 & \frac{2}{3} & 4 \end{array}\right]$$ 4. **General solution:** Let $x_4 = t$ (free variable). From $R_3$: $x_3 + \frac{2}{3} t = 4 \implies x_3 = 4 - \frac{2}{3} t$ From $R_2$: $x_2 - \frac{1}{3} t = -1 \implies x_2 = -1 + \frac{1}{3} t$ From $R_1$: $x_1 = \frac{2}{3}$ So, $$\boxed{\begin{cases}x_1 = \frac{2}{3} \\ x_2 = -1 + \frac{1}{3} t \\ x_3 = 4 - \frac{2}{3} t \\ x_4 = t \end{cases} \text{ where } t \in \mathbb{R}}$$ 5. **Particular solution with $x_1 = x_3$:** Set $x_1 = x_3$: $$\frac{2}{3} = 4 - \frac{2}{3} t$$ Solve for $t$: $$\frac{2}{3} - 4 = -\frac{2}{3} t \implies -\frac{10}{3} = -\frac{2}{3} t \implies t = 5$$ Substitute $t=5$: $$x_2 = -1 + \frac{1}{3} \times 5 = -1 + \frac{5}{3} = \frac{2}{3}$$ $$x_3 = 4 - \frac{2}{3} \times 5 = 4 - \frac{10}{3} = \frac{2}{3}$$ $$x_4 = 5$$ Particular solution: $$\boxed{\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}, 5\right)}$$ --- 6. **Vectors in $\mathbb{R}^3$:** Given points: $$A = (1,0,0), B = (0,1,0), C = (0,0,3)$$ (i) Vectors: $$\mathbf{u} = \overrightarrow{AB} = B - A = (0-1, 1-0, 0-0) = (-1, 1, 0)$$ $$\mathbf{v} = \overrightarrow{AC} = C - A = (0-1, 0-0, 3-0) = (-1, 0, 3)$$ 7. **Calculate $||\mathbf{u}||$, $\mathbf{u} \cdot \mathbf{v}$, and $\mathbf{u} \times \mathbf{v}$:** - Magnitude: $$||\mathbf{u}|| = \sqrt{(-1)^2 + 1^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}$$ - Dot product: $$\mathbf{u} \cdot \mathbf{v} = (-1)(-1) + (1)(0) + (0)(3) = 1 + 0 + 0 = 1$$ - Cross product: $$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 1 & 0 \\ -1 & 0 & 3 \end{vmatrix} = \mathbf{i}(1 \times 3 - 0 \times 0) - \mathbf{j}(-1 \times 3 - 0 \times -1) + \mathbf{k}(-1 \times 0 - 1 \times -1)$$ $$= 3\mathbf{i} - (-3)\mathbf{j} + (0 + 1)\mathbf{k} = 3\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} = (3, 3, 1)$$ 8. **Decompose $\mathbf{v}$ into $\mathbf{v}_1$ parallel to $\mathbf{u}$ and $\mathbf{v}_2$ orthogonal to $\mathbf{u}$:** - Projection formula: $$\mathbf{v}_1 = \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}||^2} \mathbf{u} = \frac{1}{2} (-1, 1, 0) = \left(-\frac{1}{2}, \frac{1}{2}, 0\right)$$ - Orthogonal component: $$\mathbf{v}_2 = \mathbf{v} - \mathbf{v}_1 = \left(-1, 0, 3\right) - \left(-\frac{1}{2}, \frac{1}{2}, 0\right) = \left(-\frac{1}{2}, -\frac{1}{2}, 3\right)$$ 9. **Area of triangle $ABC$:** Area formula: $$\text{Area} = \frac{1}{2} ||\mathbf{u} \times \mathbf{v}||$$ Calculate magnitude: $$||\mathbf{u} \times \mathbf{v}|| = \sqrt{3^2 + 3^2 + 1^2} = \sqrt{9 + 9 + 1} = \sqrt{19}$$ Area: $$\frac{1}{2} \sqrt{19}$$ --- **Final answers:** - (a)(i) RREF matrix shown above. - (a)(ii) General solution: $$\boxed{\begin{cases}x_1 = \frac{2}{3} \\ x_2 = -1 + \frac{1}{3} t \\ x_3 = 4 - \frac{2}{3} t \\ x_4 = t \end{cases}}$$ - (a)(iii) Particular solution with $x_1 = x_3$: $$\boxed{\left(\frac{2}{3}, \frac{2}{3}, \frac{2}{3}, 5\right)}$$ - (b)(i) $\mathbf{u} = (-1,1,0)$, $\mathbf{v} = (-1,0,3)$ - (b)(ii) $||\mathbf{u}|| = \sqrt{2}$, $\mathbf{u} \cdot \mathbf{v} = 1$, $\mathbf{u} \times \mathbf{v} = (3,3,1)$ - (b)(iii) $\mathbf{v}_1 = \left(-\frac{1}{2}, \frac{1}{2}, 0\right)$, $\mathbf{v}_2 = \left(-\frac{1}{2}, -\frac{1}{2}, 3\right)$ - (b)(iv) Area of triangle $ABC = \frac{1}{2} \sqrt{19}$