1. **State the problem:** We need to solve the system of linear equations given by $$\begin{bmatrix}40 & -20 & -10 \\ -20 & 50 & -20 \\ -10 & -20 & 40\end{bmatrix} \begin{bmatrix}\Theta_1 \\ \Theta_2 \\ \Theta_3\end{bmatrix} = \begin{bmatrix}150 \\ -80 \\ 120\end{bmatrix}$$ 2. **Method:** We will use LU decomposition to solve for $\Theta_1$, $\Theta_2$, and $\Theta_3$. LU decomposition factors the coefficient matrix $A$ into a lower triangular matrix $L$ and an upper triangular matrix $U$ such that $A = LU$. 3. **Step 1: Decompose matrix $A$:** $$A = \begin{bmatrix}40 & -20 & -10 \\ -20 & 50 & -20 \\ -10 & -20 & 40\end{bmatrix}$$ We find $L$ and $U$ such that: $$L = \begin{bmatrix}1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1\end{bmatrix}, \quad U = \begin{bmatrix}u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33}\end{bmatrix}$$ Calculate $u_{11} = 40$. Calculate $l_{21} = \frac{-20}{40} = -0.5$. Calculate $l_{31} = \frac{-10}{40} = -0.25$. Calculate $u_{12} = -20$. Calculate $u_{13} = -10$. Calculate $u_{22} = 50 - l_{21} \times u_{12} = 50 - (-0.5)(-20) = 50 - 10 = 40$. Calculate $u_{23} = -20 - l_{21} \times u_{13} = -20 - (-0.5)(-10) = -20 - 5 = -25$. Calculate $l_{32} = \frac{-20 - l_{31} \times u_{12}}{u_{22}} = \frac{-20 - (-0.25)(-20)}{40} = \frac{-20 - 5}{40} = \frac{-25}{40} = -0.625$. Calculate $u_{33} = 40 - l_{31} \times u_{13} - l_{32} \times u_{23} = 40 - (-0.25)(-10) - (-0.625)(-25) = 40 - 2.5 - 15.625 = 22.875$. So, $$L = \begin{bmatrix}1 & 0 & 0 \\ -0.5 & 1 & 0 \\ -0.25 & -0.625 & 1\end{bmatrix}, \quad U = \begin{bmatrix}40 & -20 & -10 \\ 0 & 40 & -25 \\ 0 & 0 & 22.875\end{bmatrix}$$ 4. **Step 2: Solve $Ly = b$ for $y$:** $$\begin{bmatrix}1 & 0 & 0 \\ -0.5 & 1 & 0 \\ -0.25 & -0.625 & 1\end{bmatrix} \begin{bmatrix}y_1 \\ y_2 \\ y_3\end{bmatrix} = \begin{bmatrix}150 \\ -80 \\ 120\end{bmatrix}$$ From first row: $y_1 = 150$. From second row: $-0.5 y_1 + y_2 = -80 \Rightarrow y_2 = -80 + 0.5 \times 150 = -80 + 75 = -5$. From third row: $-0.25 y_1 - 0.625 y_2 + y_3 = 120 \Rightarrow y_3 = 120 + 0.25 \times 150 + 0.625 \times 5 = 120 + 37.5 + 3.125 = 160.625$. 5. **Step 3: Solve $Ux = y$ for $x = \begin{bmatrix}\Theta_1 \\ \Theta_2 \\ \Theta_3\end{bmatrix}$:** $$\begin{bmatrix}40 & -20 & -10 \\ 0 & 40 & -25 \\ 0 & 0 & 22.875\end{bmatrix} \begin{bmatrix}\Theta_1 \\ \Theta_2 \\ \Theta_3\end{bmatrix} = \begin{bmatrix}150 \\ -5 \\ 160.625\end{bmatrix}$$ From third row: $22.875 \Theta_3 = 160.625 \Rightarrow \Theta_3 = \frac{160.625}{22.875} = 7.02$ (approx). From second row: $40 \Theta_2 - 25 \Theta_3 = -5 \Rightarrow 40 \Theta_2 = -5 + 25 \times 7.02 = -5 + 175.5 = 170.5 \Rightarrow \Theta_2 = \frac{170.5}{40} = 4.26$ (approx). From first row: $40 \Theta_1 - 20 \Theta_2 - 10 \Theta_3 = 150 \Rightarrow 40 \Theta_1 = 150 + 20 \times 4.26 + 10 \times 7.02 = 150 + 85.2 + 70.2 = 305.4 \Rightarrow \Theta_1 = \frac{305.4}{40} = 7.635$ (approx). **Final answer:** $$\Theta_1 \approx 7.64, \quad \Theta_2 \approx 4.26, \quad \Theta_3 \approx 7.02$$