1. **Problem Statement:** Find the least squares solution of the linear system given by $$\begin{cases} 3x_1 + x_2 - x_3 = -3 \\ x_1 + 2x_2 = -3 \\ x_2 + 2x_3 = 8 \end{cases}$$ 2. **Matrix Form:** Write the system as $A\mathbf{x} = \mathbf{b}$ where $$A = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} -3 \\ -3 \\ 8 \end{bmatrix}$$ 3. **Least Squares Solution:** Since the system may be inconsistent, the least squares solution minimizes $\|A\mathbf{x} - \mathbf{b}\|^2$. 4. **Normal Equations:** Solve $$A^T A \mathbf{x} = A^T \mathbf{b}$$ Calculate $$A^T = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 2 \end{bmatrix}$$ 5. Compute $$A^T A = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 & -1 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 10 & 5 & -3 \\ 5 & 6 & 2 \\ -3 & 2 & 5 \end{bmatrix}$$ 6. Compute $$A^T \mathbf{b} = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 2 & 1 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} -3 \\ -3 \\ 8 \end{bmatrix} = \begin{bmatrix} -12 \\ -1 \\ 13 \end{bmatrix}$$ 7. **Solve the system:** $$\begin{bmatrix} 10 & 5 & -3 \\ 5 & 6 & 2 \\ -3 & 2 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -12 \\ -1 \\ 13 \end{bmatrix}$$ Using substitution or matrix methods: - Multiply first equation by 1: $10x_1 + 5x_2 - 3x_3 = -12$ - Multiply second equation by 2: $10x_1 + 12x_2 + 4x_3 = -2$ - Subtract first from second: $(12-5)x_2 + (4+3)x_3 = (-2+12)$ gives $$7x_2 + 7x_3 = 10 \implies x_2 + x_3 = \frac{10}{7}$$ - From third equation: $-3x_1 + 2x_2 + 5x_3 = 13$ - Express $x_1$ from first equation: $$10x_1 = -12 - 5x_2 + 3x_3 \implies x_1 = \frac{-12 - 5x_2 + 3x_3}{10}$$ - Substitute $x_1$ into third equation: $$-3 \times \frac{-12 - 5x_2 + 3x_3}{10} + 2x_2 + 5x_3 = 13$$ Multiply both sides by 10: $$36 + 15x_2 - 9x_3 + 20x_2 + 50x_3 = 130$$ Simplify: $$35x_2 + 41x_3 = 94$$ - Recall from above $x_2 = \frac{10}{7} - x_3$, substitute: $$35 \left( \frac{10}{7} - x_3 \right) + 41x_3 = 94$$ $$50 - 35x_3 + 41x_3 = 94$$ $$6x_3 = 44 \implies x_3 = \frac{44}{6} = \frac{22}{3}$$ - Then $$x_2 = \frac{10}{7} - \frac{22}{3} = \frac{30 - 154}{21} = -\frac{124}{21}$$ - Finally, $$x_1 = \frac{-12 - 5 \times (-\frac{124}{21}) + 3 \times \frac{22}{3}}{10} = \frac{-12 + \frac{620}{21} + 22}{10} = \frac{-12 + 22 + \frac{620}{21}}{10} = \frac{10 + \frac{620}{21}}{10} = \frac{\frac{210}{21} + 620}{210} = \frac{830}{210} = \frac{83}{21}$$ 8. **Least Squares Solution:** $$\boxed{\mathbf{x} = \left( \frac{83}{21}, -\frac{124}{21}, \frac{22}{3} \right)}$$ 9. **Orthogonal Projection:** Compute $$\hat{\mathbf{b}} = A \mathbf{x} = \begin{bmatrix} 3 & 1 & -1 \\ 1 & 2 & 0 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} \frac{83}{21} \\ -\frac{124}{21} \\ \frac{22}{3} \end{bmatrix} = \begin{bmatrix} 3 \times \frac{83}{21} + 1 \times \left(-\frac{124}{21}\right) - 1 \times \frac{22}{3} \\ 1 \times \frac{83}{21} + 2 \times \left(-\frac{124}{21}\right) + 0 \\ 0 + 1 \times \left(-\frac{124}{21}\right) + 2 \times \frac{22}{3} \end{bmatrix}$$ Calculate each: - First component: $$\frac{249}{21} - \frac{124}{21} - \frac{154}{21} = \frac{249 - 124 - 154}{21} = \frac{-29}{21}$$ - Second component: $$\frac{83}{21} - \frac{248}{21} = -\frac{165}{21} = -\frac{55}{7}$$ - Third component: $$-\frac{124}{21} + \frac{308}{21} = \frac{184}{21}$$ So, $$\hat{\mathbf{b}} = \begin{bmatrix} -\frac{29}{21} \\ -\frac{55}{7} \\ \frac{184}{21} \end{bmatrix}$$ 10. **Least Squares Error:** $$\| \mathbf{b} - \hat{\mathbf{b}} \|^2 = \left\| \begin{bmatrix} -3 \\ -3 \\ 8 \end{bmatrix} - \begin{bmatrix} -\frac{29}{21} \\ -\frac{55}{7} \\ \frac{184}{21} \end{bmatrix} \right\|^2 = \left\| \begin{bmatrix} -3 + \frac{29}{21} \\ -3 + \frac{55}{7} \\ 8 - \frac{184}{21} \end{bmatrix} \right\|^2$$ Calculate each difference: - $-3 + \frac{29}{21} = -\frac{63}{21} + \frac{29}{21} = -\frac{34}{21}$ - $-3 + \frac{55}{7} = -\frac{21}{7} + \frac{55}{7} = \frac{34}{7}$ - $8 - \frac{184}{21} = \frac{168}{21} - \frac{184}{21} = -\frac{16}{21}$ Square and sum: $$\left(-\frac{34}{21}\right)^2 + \left(\frac{34}{7}\right)^2 + \left(-\frac{16}{21}\right)^2 = \frac{1156}{441} + \frac{1156}{49} + \frac{256}{441} = \frac{1156}{441} + \frac{1156 \times 9}{441} + \frac{256}{441} = \frac{1156 + 10404 + 256}{441} = \frac{11916}{441} = \frac{1324}{49} \approx 27.02$$ **Final answers:** - Least squares solution: $$\boxed{\mathbf{x} = \left( \frac{83}{21}, -\frac{124}{21}, \frac{22}{3} \right)}$$ - Orthogonal projection of $\mathbf{b}$ on column space of $A$: $$\boxed{\hat{\mathbf{b}} = \left( -\frac{29}{21}, -\frac{55}{7}, \frac{184}{21} \right)}$$ - Least squares error: $$\boxed{\| \mathbf{b} - \hat{\mathbf{b}} \|^2 = \frac{1324}{49} \approx 27.02}$$