1. **State the problem:** Solve the system of linear equations given by $$\begin{pmatrix} 3 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \\ -10 \end{pmatrix}$$ using the Cholesky decomposition method. 2. **Recall the Cholesky decomposition:** For a symmetric positive definite matrix $A$, we can write $$A = LL^T$$ where $L$ is a lower triangular matrix. 3. **Check symmetry:** The matrix $$A = \begin{pmatrix} 3 & 2 & 1 \\ 2 & 3 & 2 \\ 1 & 2 & 2 \end{pmatrix}$$ is symmetric since $A = A^T$. 4. **Find $L$ such that $A = LL^T$:** Let $$L = \begin{pmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{pmatrix}$$ We have the equations from $A = LL^T$: - $l_{11}^2 = 3$ - $l_{21} l_{11} = 2$ - $l_{31} l_{11} = 1$ - $l_{21}^2 + l_{22}^2 = 3$ - $l_{31} l_{21} + l_{32} l_{22} = 2$ - $l_{31}^2 + l_{32}^2 + l_{33}^2 = 2$ 5. **Calculate $l_{11}$:** $$l_{11} = \sqrt{3}$$ 6. **Calculate $l_{21}$:** $$l_{21} = \frac{2}{l_{11}} = \frac{2}{\sqrt{3}}$$ 7. **Calculate $l_{31}$:** $$l_{31} = \frac{1}{l_{11}} = \frac{1}{\sqrt{3}}$$ 8. **Calculate $l_{22}$:** $$l_{22} = \sqrt{3 - l_{21}^2} = \sqrt{3 - \left(\frac{2}{\sqrt{3}}\right)^2} = \sqrt{3 - \frac{4}{3}} = \sqrt{\frac{5}{3}}$$ 9. **Calculate $l_{32}$:** $$l_{32} = \frac{2 - l_{31} l_{21}}{l_{22}} = \frac{2 - \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}}}{\sqrt{\frac{5}{3}}} = \frac{2 - \frac{2}{3}}{\sqrt{\frac{5}{3}}} = \frac{\frac{4}{3}}{\sqrt{\frac{5}{3}}} = \frac{4}{3} \cdot \sqrt{\frac{3}{5}} = \frac{4}{3} \cdot \frac{\sqrt{15}}{5} = \frac{4 \sqrt{15}}{15}$$ 10. **Calculate $l_{33}$:** $$l_{33} = \sqrt{2 - l_{31}^2 - l_{32}^2} = \sqrt{2 - \left(\frac{1}{\sqrt{3}}\right)^2 - \left(\frac{4 \sqrt{15}}{15}\right)^2} = \sqrt{2 - \frac{1}{3} - \frac{16 \cdot 15}{225}} = \sqrt{2 - \frac{1}{3} - \frac{16}{15}} = \sqrt{\frac{30}{15} - \frac{5}{15} - \frac{16}{15}} = \sqrt{\frac{9}{15}} = \sqrt{\frac{3}{5}}$$ 11. **Write $L$ explicitly:** $$L = \begin{pmatrix} \sqrt{3} & 0 & 0 \\ \frac{2}{\sqrt{3}} & \sqrt{\frac{5}{3}} & 0 \\ \frac{1}{\sqrt{3}} & \frac{4 \sqrt{15}}{15} & \sqrt{\frac{3}{5}} \end{pmatrix}$$ 12. **Solve $Ly = b$ where $b = \begin{pmatrix} 5 \\ 6 \\ -10 \end{pmatrix}$:** - $\sqrt{3} y_1 = 5 \implies y_1 = \frac{5}{\sqrt{3}}$ - $\frac{2}{\sqrt{3}} y_1 + \sqrt{\frac{5}{3}} y_2 = 6$ - $\frac{1}{\sqrt{3}} y_1 + \frac{4 \sqrt{15}}{15} y_2 + \sqrt{\frac{3}{5}} y_3 = -10$ 13. **Calculate $y_2$:** $$\frac{2}{\sqrt{3}} \cdot \frac{5}{\sqrt{3}} + \sqrt{\frac{5}{3}} y_2 = 6 \implies \frac{10}{3} + \sqrt{\frac{5}{3}} y_2 = 6$$ $$\sqrt{\frac{5}{3}} y_2 = 6 - \frac{10}{3} = \frac{18}{3} - \frac{10}{3} = \frac{8}{3}$$ $$y_2 = \frac{8}{3} \div \sqrt{\frac{5}{3}} = \frac{8}{3} \cdot \sqrt{\frac{3}{5}} = \frac{8}{3} \cdot \frac{\sqrt{15}}{5} = \frac{8 \sqrt{15}}{15}$$ 14. **Calculate $y_3$:** $$\frac{1}{\sqrt{3}} \cdot \frac{5}{\sqrt{3}} + \frac{4 \sqrt{15}}{15} \cdot \frac{8 \sqrt{15}}{15} + \sqrt{\frac{3}{5}} y_3 = -10$$ Calculate each term: $$\frac{5}{3} + \frac{4 \sqrt{15}}{15} \cdot \frac{8 \sqrt{15}}{15} + \sqrt{\frac{3}{5}} y_3 = -10$$ $$\frac{5}{3} + \frac{4 \cdot 8 \cdot 15}{15 \cdot 15} + \sqrt{\frac{3}{5}} y_3 = -10$$ $$\frac{5}{3} + \frac{480}{225} + \sqrt{\frac{3}{5}} y_3 = -10$$ Simplify $\frac{480}{225} = \frac{64}{30} = \frac{32}{15}$: $$\frac{5}{3} + \frac{32}{15} + \sqrt{\frac{3}{5}} y_3 = -10$$ Convert to common denominator 15: $$\frac{25}{15} + \frac{32}{15} + \sqrt{\frac{3}{5}} y_3 = -10$$ $$\frac{57}{15} + \sqrt{\frac{3}{5}} y_3 = -10$$ $$\sqrt{\frac{3}{5}} y_3 = -10 - \frac{57}{15} = -\frac{150}{15} - \frac{57}{15} = -\frac{207}{15}$$ $$y_3 = -\frac{207}{15} \div \sqrt{\frac{3}{5}} = -\frac{207}{15} \cdot \sqrt{\frac{5}{3}} = -\frac{207}{15} \cdot \frac{\sqrt{15}}{3} = -\frac{207 \sqrt{15}}{45} = -\frac{69 \sqrt{15}}{15}$$ 15. **Solve $L^T x = y$:** $$L^T = \begin{pmatrix} \sqrt{3} & \frac{2}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ 0 & \sqrt{\frac{5}{3}} & \frac{4 \sqrt{15}}{15} \\ 0 & 0 & \sqrt{\frac{3}{5}} \end{pmatrix}$$ The system is: - $\sqrt{3} x_1 + \frac{2}{\sqrt{3}} x_2 + \frac{1}{\sqrt{3}} x_3 = y_1$ - $\sqrt{\frac{5}{3}} x_2 + \frac{4 \sqrt{15}}{15} x_3 = y_2$ - $\sqrt{\frac{3}{5}} x_3 = y_3$ 16. **Calculate $x_3$:** $$x_3 = \frac{y_3}{\sqrt{\frac{3}{5}}} = y_3 \cdot \sqrt{\frac{5}{3}} = -\frac{69 \sqrt{15}}{15} \cdot \sqrt{\frac{5}{3}} = -\frac{69 \sqrt{15}}{15} \cdot \frac{\sqrt{15}}{3} = -\frac{69 \cdot 15}{15 \cdot 3} = -\frac{69}{3} = -23$$ 17. **Calculate $x_2$:** $$\sqrt{\frac{5}{3}} x_2 + \frac{4 \sqrt{15}}{15} (-23) = \frac{8 \sqrt{15}}{15}$$ $$\sqrt{\frac{5}{3}} x_2 - \frac{92 \sqrt{15}}{15} = \frac{8 \sqrt{15}}{15}$$ $$\sqrt{\frac{5}{3}} x_2 = \frac{8 \sqrt{15}}{15} + \frac{92 \sqrt{15}}{15} = \frac{100 \sqrt{15}}{15} = \frac{20 \sqrt{15}}{3}$$ $$x_2 = \frac{20 \sqrt{15}}{3} \div \sqrt{\frac{5}{3}} = \frac{20 \sqrt{15}}{3} \cdot \sqrt{\frac{3}{5}} = \frac{20 \sqrt{15}}{3} \cdot \frac{\sqrt{15}}{5} = \frac{20 \cdot 15}{3 \cdot 5} = \frac{300}{15} = 20$$ 18. **Calculate $x_1$:** $$\sqrt{3} x_1 + \frac{2}{\sqrt{3}} (20) + \frac{1}{\sqrt{3}} (-23) = \frac{5}{\sqrt{3}}$$ $$\sqrt{3} x_1 + \frac{40}{\sqrt{3}} - \frac{23}{\sqrt{3}} = \frac{5}{\sqrt{3}}$$ $$\sqrt{3} x_1 + \frac{17}{\sqrt{3}} = \frac{5}{\sqrt{3}}$$ $$\sqrt{3} x_1 = \frac{5}{\sqrt{3}} - \frac{17}{\sqrt{3}} = -\frac{12}{\sqrt{3}}$$ $$x_1 = -\frac{12}{\sqrt{3}} \div \sqrt{3} = -\frac{12}{3} = -4$$ **Final solution:** $$\boxed{\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} -4 \\ 20 \\ -23 \end{pmatrix}}$$