1. **Problem Statement:** Solve the linear system using LU decomposition: $$\begin{cases} 2x + 8y = -2 \\ -x - y = -2 \end{cases}$$ 2. **LU Decomposition Concept:** LU decomposition factors a matrix $A$ into a lower triangular matrix $L$ and an upper triangular matrix $U$ such that $A = LU$. 3. **Matrix Setup:** Coefficient matrix: $$A = \begin{bmatrix} 2 & 8 \\ -1 & -1 \end{bmatrix}$$ Right-hand side vector: $$b = \begin{bmatrix} -2 \\ -2 \end{bmatrix}$$ 4. **Decompose $A$ into $L$ and $U$:** Assume: $$L = \begin{bmatrix} 1 & 0 \\ l_{21} & 1 \end{bmatrix}, \quad U = \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix}$$ From $A = LU$: - $u_{11} = 2$ - $u_{12} = 8$ - $l_{21} \times u_{11} = -1 \implies l_{21} = \frac{-1}{2} = -0.5$ - $l_{21} \times u_{12} + u_{22} = -1 \implies (-0.5)(8) + u_{22} = -1 \implies -4 + u_{22} = -1 \implies u_{22} = 3$ So: $$L = \begin{bmatrix} 1 & 0 \\ -0.5 & 1 \end{bmatrix}, \quad U = \begin{bmatrix} 2 & 8 \\ 0 & 3 \end{bmatrix}$$ 5. **Solve $Ly = b$ for $y$:** $$\begin{bmatrix} 1 & 0 \\ -0.5 & 1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \end{bmatrix} = \begin{bmatrix} -2 \\ -2 \end{bmatrix}$$ - From first row: $y_1 = -2$ - From second row: $-0.5 y_1 + y_2 = -2 \implies -0.5(-2) + y_2 = -2 \implies 1 + y_2 = -2 \implies y_2 = -3$ 6. **Solve $Ux = y$ for $x$:** $$\begin{bmatrix} 2 & 8 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -2 \\ -3 \end{bmatrix}$$ - From second row: $3y = -3 \implies y = -1$ - From first row: $2x + 8(-1) = -2 \implies 2x - 8 = -2 \implies 2x = 6 \implies x = 3$ 7. **Final solution:** $$\boxed{x = 3, \quad y = -1}$$ This completes the LU decomposition and solution of the system.