1. **Problem statement:** Find the inverse of the matrix $$A = \begin{bmatrix} 5 & 8 & 1 \\ 0 & 2 & 1 \\ 4 & 3 & -1 \end{bmatrix}$$ using LU decomposition with $$l_{11} = l_{22} = l_{33} = 1$$. 2. **LU decomposition method:** We decompose $$A$$ into $$L$$ (lower triangular) and $$U$$ (upper triangular) matrices such that $$A = LU$$, where $$L$$ has ones on the diagonal. 3. **Set up matrices:** $$L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$$ 4. **Find $$U$$ first row:** $$u_{11} = 5, \quad u_{12} = 8, \quad u_{13} = 1$$ 5. **Find $$L$$ first column:** From $$A_{21} = l_{21} u_{11} \Rightarrow 0 = l_{21} \times 5 \Rightarrow l_{21} = 0$$ From $$A_{31} = l_{31} u_{11} \Rightarrow 4 = l_{31} \times 5 \Rightarrow l_{31} = \frac{4}{5}$$ 6. **Find $$U$$ second row:** $$A_{22} = l_{21} u_{12} + u_{22} \Rightarrow 2 = 0 \times 8 + u_{22} \Rightarrow u_{22} = 2$$ $$A_{23} = l_{21} u_{13} + u_{23} \Rightarrow 1 = 0 \times 1 + u_{23} \Rightarrow u_{23} = 1$$ 7. **Find $$L$$ second column:** $$A_{32} = l_{31} u_{12} + l_{32} u_{22} \Rightarrow 3 = \frac{4}{5} \times 8 + l_{32} \times 2$$ Calculate $$\frac{4}{5} \times 8 = \frac{32}{5} = 6.4$$ So, $$3 = 6.4 + 2 l_{32} \Rightarrow 2 l_{32} = 3 - 6.4 = -3.4 \Rightarrow l_{32} = -1.7$$ 8. **Find $$U$$ third row:** $$A_{33} = l_{31} u_{13} + l_{32} u_{23} + u_{33} \Rightarrow -1 = \frac{4}{5} \times 1 + (-1.7) \times 1 + u_{33}$$ Calculate $$\frac{4}{5} = 0.8$$ So, $$-1 = 0.8 - 1.7 + u_{33} \Rightarrow u_{33} = -1 - 0.8 + 1.7 = -0.1$$ 9. **Summary:** $$L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \frac{4}{5} & -1.7 & 1 \end{bmatrix}, \quad U = \begin{bmatrix} 5 & 8 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -0.1 \end{bmatrix}$$ 10. **Find inverse:** Solve $$AX = I$$ by solving $$LY = I$$ and then $$UX = Y$$ for each column of identity matrix $$I$$. 11. **Solve for first column:** $$LY = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ From $$L$$: $$y_1 = 1$$ $$0 = 0 \times y_1 + y_2 \Rightarrow y_2 = 0$$ $$0 = \frac{4}{5} y_1 -1.7 y_2 + y_3 \Rightarrow y_3 = -\frac{4}{5} = -0.8$$ 12. **Solve $$UX = Y$$ for first column:** $$U \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ -0.8 \end{bmatrix}$$ From bottom up: $$-0.1 x_3 = -0.8 \Rightarrow x_3 = \frac{-0.8}{-0.1} = 8$$ $$2 x_2 + 1 \times 8 = 0 \Rightarrow 2 x_2 = -8 \Rightarrow x_2 = -4$$ $$5 x_1 + 8 x_2 + 1 x_3 = 1 \Rightarrow 5 x_1 + 8(-4) + 8 = 1 \Rightarrow 5 x_1 - 32 + 8 = 1 \Rightarrow 5 x_1 - 24 = 1 \Rightarrow 5 x_1 = 25 \Rightarrow x_1 = 5$$ 13. **First column of inverse:** $$\begin{bmatrix} 5 \\ -4 \\ 8 \end{bmatrix}$$ 14. **Repeat steps 11-13 for second and third columns of identity matrix:** Second column $$I_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$ Solve $$LY = I_2$$: $$y_1 = 0$$ $$y_2 = 1$$ $$y_3 = -\frac{4}{5} y_1 + 1.7 y_2 = 0 + 1.7 = 1.7$$ Solve $$UX = Y$$: $$-0.1 x_3 = 1.7 \Rightarrow x_3 = -17$$ $$2 x_2 + 1 x_3 = 1 \Rightarrow 2 x_2 - 17 = 1 \Rightarrow 2 x_2 = 18 \Rightarrow x_2 = 9$$ $$5 x_1 + 8 x_2 + 1 x_3 = 0 \Rightarrow 5 x_1 + 72 - 17 = 0 \Rightarrow 5 x_1 + 55 = 0 \Rightarrow 5 x_1 = -55 \Rightarrow x_1 = -11$$ Second column inverse: $$\begin{bmatrix} -11 \\ 9 \\ -17 \end{bmatrix}$$ Third column $$I_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ Solve $$LY = I_3$$: $$y_1 = 0$$ $$y_2 = 0$$ $$y_3 = 1$$ Solve $$UX = Y$$: $$-0.1 x_3 = 1 \Rightarrow x_3 = -10$$ $$2 x_2 + 1 x_3 = 0 \Rightarrow 2 x_2 - 10 = 0 \Rightarrow 2 x_2 = 10 \Rightarrow x_2 = 5$$ $$5 x_1 + 8 x_2 + 1 x_3 = 0 \Rightarrow 5 x_1 + 40 - 10 = 0 \Rightarrow 5 x_1 + 30 = 0 \Rightarrow 5 x_1 = -30 \Rightarrow x_1 = -6$$ Third column inverse: $$\begin{bmatrix} -6 \\ 5 \\ -10 \end{bmatrix}$$ 15. **Final inverse matrix:** $$A^{-1} = \begin{bmatrix} 5 & -11 & -6 \\ -4 & 9 & 5 \\ 8 & -17 & -10 \end{bmatrix}$$