1. **State the problem:** We want to solve the system $Ax=\bar{b}$ where $$A=\begin{bmatrix}4 & 2 & 0 \\ -8 & -2 & -7 \\ 13 & 11 & 21 \\ \end{bmatrix}, \quad \bar{b}=\begin{bmatrix}1 \\ 17 \\ 17 \\ \end{bmatrix}$$ Given the LU factorization: $$A = L U = \begin{bmatrix}1 & 5 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \\ \end{bmatrix} \cdot \begin{bmatrix}4 & 0 & -2 \\ 3 & 0 & 17 \\ -2 & 0 & 7 \\ \end{bmatrix}$$ (Note: The factorization matrices are as given, but typically $L$ is lower triangular and $U$ upper triangular. Here, we use the provided matrices as $L$ and $U$.) 2. **Use the LU factorization to solve $Ax=\bar{b}$:** Since $Ax=LUx=\bar{b}$, let $Ux=y$. Then we solve two systems: - First, solve $Ly=\bar{b}$ for $y$. - Then solve $Ux=y$ for $x$. 3. **Solve $Ly=\bar{b}$:** $$\begin{bmatrix}1 & 5 & -2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix}y_1 \\ y_2 \\ y_3 \\ \end{bmatrix} = \begin{bmatrix}1 \\ 17 \\ 17 \\ \end{bmatrix}$$ From the third row: $$1 \cdot y_3 = 17 \implies y_3 = 17$$ From the second row: $$1 \cdot y_2 + 3 \cdot y_3 = 17 \implies y_2 + 3 \times 17 = 17 \implies y_2 + 51 = 17 \implies y_2 = 17 - 51 = -34$$ From the first row: $$1 \cdot y_1 + 5 \cdot y_2 - 2 \cdot y_3 = 1 \implies y_1 + 5(-34) - 2(17) = 1 \implies y_1 - 170 - 34 = 1 \implies y_1 - 204 = 1 \implies y_1 = 205$$ So, $$y = \begin{bmatrix}205 \\ -34 \\ 17 \end{bmatrix}$$ 4. **Solve $Ux = y$:** $$\begin{bmatrix}4 & 0 & -2 \\ 3 & 0 & 17 \\ -2 & 0 & 7 \\ \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix}205 \\ -34 \\ 17 \\ \end{bmatrix}$$ From the system: - First row: $$4x_1 - 2x_3 = 205$$ - Second row: $$3x_1 + 17x_3 = -34$$ - Third row: $$-2x_1 + 7x_3 = 17$$ Note that $x_2$ does not appear in these equations, so it is a free variable. We can set $x_2 = t$ for some parameter $t$. 5. **Solve for $x_1$ and $x_3$ using the first and second equations:** From the first equation: $$4x_1 - 2x_3 = 205 \implies 2x_1 - x_3 = 102.5$$ From the second equation: $$3x_1 + 17x_3 = -34$$ Multiply the first modified equation by 17: $$34x_1 - 17x_3 = 1742.5$$ Add to the second equation: $$(3x_1 + 17x_3) + (34x_1 - 17x_3) = -34 + 1742.5 \implies 37x_1 = 1708.5 \implies x_1 = \frac{1708.5}{37} = 46.162162...$$ Substitute $x_1$ back into $2x_1 - x_3 = 102.5$: $$2(46.162162) - x_3 = 102.5 \implies 92.324324 - x_3 = 102.5 \implies x_3 = 92.324324 - 102.5 = -10.175676$$ 6. **Check the third equation for consistency:** $$-2x_1 + 7x_3 = -2(46.162162) + 7(-10.175676) = -92.324324 - 71.229732 = -163.554056$$ This does not equal 17, so the system is inconsistent with the given LU factorization or matrices. **Conclusion:** The given LU factorization or matrices appear inconsistent with the system $Ax=\bar{b}$. Normally, $L$ should be lower triangular and $U$ upper triangular, but the provided matrices do not satisfy this. Therefore, the system cannot be solved as stated with the given LU factorization. **If we ignore the inconsistency and treat $x_2$ as free, the solution is:** $$x = \begin{bmatrix}46.162 \\ t \\ -10.176 \end{bmatrix}, \quad t \in \mathbb{R}$$