1. **Problem Statement:** Obtain the LU factorization of the matrix $$A = \begin{pmatrix} 1 & 1 & 1 \\ 4 & 3 & -1 \\ 3 & 5 & 3 \end{pmatrix}$$ and use it to solve the system $Ax = b$ where $$b = \begin{pmatrix} 1 \\ 6 \\ 4 \end{pmatrix}$$ 2. **LU Factorization:** We want to write $A = LU$ where $L$ is lower triangular with 1's on the diagonal and $U$ is upper triangular. 3. **Step 1: Initialize $L$ and $U$:** $$L = \begin{pmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{pmatrix}$$ 4. **Step 2: Compute $U$ first row and $L$ first column:** - $u_{11} = 1$ - $u_{12} = 1$ - $u_{13} = 1$ - $l_{21} = \frac{4}{1} = 4$ - $l_{31} = \frac{3}{1} = 3$ 5. **Step 3: Compute $U$ second row:** $$u_{22} = 3 - l_{21} \times u_{12} = 3 - 4 \times 1 = 3 - 4 = -1$$ $$u_{23} = -1 - l_{21} \times u_{13} = -1 - 4 \times 1 = -1 - 4 = -5$$ 6. **Step 4: Compute $L$ third row, second column:** $$l_{32} = \frac{5 - l_{31} \times u_{12}}{u_{22}} = \frac{5 - 3 \times 1}{-1} = \frac{2}{-1} = -2$$ 7. **Step 5: Compute $U$ third row, third column:** $$u_{33} = 3 - l_{31} \times u_{13} - l_{32} \times u_{23} = 3 - 3 \times 1 - (-2) \times (-5) = 3 - 3 - 10 = -10$$ 8. **Final LU matrices:** $$L = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{pmatrix}$$ 9. **Step 6: Solve $Ly = b$ for $y$:** $$\begin{cases} y_1 = 1 \\ y_2 = 6 - 4y_1 = 6 - 4 \times 1 = 2 \\ y_3 = 4 - 3y_1 + 2y_2 = 4 - 3 \times 1 + 2 \times 2 = 4 - 3 + 4 = 5 \end{cases}$$ 10. **Step 7: Solve $Ux = y$ for $x$:** $$\begin{cases} -10x_3 = 5 \implies x_3 = \frac{5}{-10} = -\frac{1}{2} \\ -1x_2 - 5x_3 = 2 \implies -x_2 - 5 \times (-\frac{1}{2}) = 2 \implies -x_2 + \frac{5}{2} = 2 \implies -x_2 = 2 - \frac{5}{2} = -\frac{1}{2} \implies x_2 = \frac{1}{2} \\ 1x_1 + 1x_2 + 1x_3 = 1 \implies x_1 + \frac{1}{2} - \frac{1}{2} = 1 \implies x_1 = 1 \end{cases}$$ 11. **Final solution:** $$x = \begin{pmatrix} 1 \\ \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix}$$ **Answer:** The LU factorization is $$L = \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 3 & -2 & 1 \end{pmatrix}, U = \begin{pmatrix} 1 & 1 & 1 \\ 0 & -1 & -5 \\ 0 & 0 & -10 \end{pmatrix}$$ and the solution to $Ax = b$ is $$x = \begin{pmatrix} 1 \\ 0.5 \\ -0.5 \end{pmatrix}$$