1. **Problem statement:** We are given the system of linear equations: $$\begin{cases} 4x + y + z = 6 \\ x + 3y + z = 5 \\ x + y + 2z = 4 \end{cases}$$ We want to compute the Cholesky decomposition of the coefficient matrix and use it to solve the system. 2. **Coefficient matrix and vector:** The coefficient matrix is $$A = \begin{bmatrix} 4 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$$ and the right-hand side vector is $$b = \begin{bmatrix} 6 \\ 5 \\ 4 \end{bmatrix}$$ 3. **Cholesky decomposition:** Cholesky decomposition factorizes a symmetric positive definite matrix $A$ into $$A = LL^T$$ where $L$ is a lower triangular matrix. 4. **Compute $L$ step-by-step:** Let $$L = \begin{bmatrix} l_{11} & 0 & 0 \\ l_{21} & l_{22} & 0 \\ l_{31} & l_{32} & l_{33} \end{bmatrix}$$ From $A = LL^T$, equate elements: - $l_{11}^2 = 4 \implies l_{11} = 2$ - $l_{21} l_{11} = 1 \implies l_{21} = \frac{1}{2} = 0.5$ - $l_{31} l_{11} = 1 \implies l_{31} = \frac{1}{2} = 0.5$ - $l_{21}^2 + l_{22}^2 = 3 \implies (0.5)^2 + l_{22}^2 = 3 \implies l_{22}^2 = 3 - 0.25 = 2.75 \implies l_{22} = \sqrt{2.75} \approx 1.6583$ - $l_{21} l_{31} + l_{22} l_{32} = 1 \implies (0.5)(0.5) + 1.6583 l_{32} = 1 \implies 0.25 + 1.6583 l_{32} = 1 \implies l_{32} = \frac{0.75}{1.6583} \approx 0.4523$ - $l_{31}^2 + l_{32}^2 + l_{33}^2 = 2 \implies (0.5)^2 + (0.4523)^2 + l_{33}^2 = 2 \implies 0.25 + 0.2046 + l_{33}^2 = 2 \implies l_{33}^2 = 2 - 0.4546 = 1.5454 \implies l_{33} = \sqrt{1.5454} \approx 1.2431$ So, $$L \approx \begin{bmatrix} 2 & 0 & 0 \\ 0.5 & 1.6583 & 0 \\ 0.5 & 0.4523 & 1.2431 \end{bmatrix}$$ 5. **Solve $Ly = b$ for $y$ (forward substitution):** $$\begin{cases} 2 y_1 = 6 \implies y_1 = 3 \\ 0.5 y_1 + 1.6583 y_2 = 5 \implies 0.5(3) + 1.6583 y_2 = 5 \implies 1.5 + 1.6583 y_2 = 5 \implies y_2 = \frac{3.5}{1.6583} \approx 2.109 \\ 0.5 y_1 + 0.4523 y_2 + 1.2431 y_3 = 4 \implies 0.5(3) + 0.4523(2.109) + 1.2431 y_3 = 4 \implies 1.5 + 0.954 + 1.2431 y_3 = 4 \implies 1.2431 y_3 = 4 - 2.454 = 1.546 \implies y_3 = \frac{1.546}{1.2431} \approx 1.243 \end{cases}$$ 6. **Solve $L^T x = y$ for $x$ (back substitution):** $$L^T = \begin{bmatrix} 2 & 0.5 & 0.5 \\ 0 & 1.6583 & 0.4523 \\ 0 & 0 & 1.2431 \end{bmatrix}$$ Equations: $$\begin{cases} 1.2431 x_3 = 1.243 \implies x_3 = 1 \\ 1.6583 x_2 + 0.4523 x_3 = 2.109 \implies 1.6583 x_2 + 0.4523(1) = 2.109 \implies 1.6583 x_2 = 1.6567 \implies x_2 = 1 \\ 2 x_1 + 0.5 x_2 + 0.5 x_3 = 3 \implies 2 x_1 + 0.5(1) + 0.5(1) = 3 \implies 2 x_1 + 1 = 3 \implies 2 x_1 = 2 \implies x_1 = 1 \end{cases}$$ 7. **Final solution:** $$\boxed{\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}}$$