1. **Problem 4: Solve the system using Jacobi's method with initial guess $x^{(0)} = 0$ up to two iterations.** Given system: $$10x_1 - x_2 = 25$$ $$x_1 + 8x_2 = 43$$ 2. **Jacobi's method formula:** For a system $Ax = b$, the iterative formula for each variable is: $$x_i^{(k+1)} = \frac{1}{a_{ii}} \left(b_i - \sum_{j \neq i} a_{ij} x_j^{(k)}\right)$$ 3. **Rewrite equations to isolate variables:** $$x_1 = \frac{25 + x_2}{10}$$ $$x_2 = \frac{43 - x_1}{8}$$ 4. **Initial guess:** $$x^{(0)} = (0,0)$$ 5. **Iteration 1:** Calculate $x_1^{(1)}$: $$x_1^{(1)} = \frac{25 + 0}{10} = 2.5$$ Calculate $x_2^{(1)}$: $$x_2^{(1)} = \frac{43 - 0}{8} = 5.375$$ 6. **Iteration 2:** Calculate $x_1^{(2)}$: $$x_1^{(2)} = \frac{25 + 5.375}{10} = \frac{30.375}{10} = 3.0375$$ Calculate $x_2^{(2)}$: $$x_2^{(2)} = \frac{43 - 2.5}{8} = \frac{40.5}{8} = 5.0625$$ 7. **Final approximate solution after two iterations:** $$x^{(2)} = (3.0375, 5.0625)$$ --- 1. **Problem 1: Find the standard matrix $A$ representing the linear operator $L: \mathbb{R}^2 \to \mathbb{R}^2$ defined by** $$L(x,y) = \begin{pmatrix} x + 2y \\ x + y \end{pmatrix}$$ 2. **Recall:** The standard matrix $A$ is formed by applying $L$ to the standard basis vectors $e_1 = (1,0)$ and $e_2 = (0,1)$: $$A = \begin{pmatrix} L(e_1) & L(e_2) \end{pmatrix}$$ 3. **Calculate:** $$L(e_1) = L(1,0) = \begin{pmatrix} 1 + 2\cdot0 \\ 1 + 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ $$L(e_2) = L(0,1) = \begin{pmatrix} 0 + 2\cdot1 \\ 0 + 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$ 4. **Thus, the matrix is:** $$A = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$$ --- 1. **Problem 2: Determine if matrices $A$ and $B$ commute, i.e., if $AB = BA$.** Given: $$A = \begin{pmatrix} 3 & 2 \\ -1 & 3 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & -2 \\ 2 & 1 \end{pmatrix}$$ 2. **Calculate $AB$:** $$AB = \begin{pmatrix} 3 & 2 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} -1 & -2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 3\cdot(-1) + 2\cdot2 & 3\cdot(-2) + 2\cdot1 \\ -1\cdot(-1) + 3\cdot2 & -1\cdot(-2) + 3\cdot1 \end{pmatrix} = \begin{pmatrix} -3 + 4 & -6 + 2 \\ 1 + 6 & 2 + 3 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 7 & 5 \end{pmatrix}$$ 3. **Calculate $BA$:** $$BA = \begin{pmatrix} -1 & -2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} -1\cdot3 + (-2)\cdot(-1) & -1\cdot2 + (-2)\cdot3 \\ 2\cdot3 + 1\cdot(-1) & 2\cdot2 + 1\cdot3 \end{pmatrix} = \begin{pmatrix} -3 + 2 & -2 - 6 \\ 6 - 1 & 4 + 3 \end{pmatrix} = \begin{pmatrix} -1 & -8 \\ 5 & 7 \end{pmatrix}$$ 4. **Since $AB \neq BA$, the matrices do not commute.** --- 1. **Problem 3: Find an LU decomposition of matrix** $$A = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$$ 2. **LU decomposition expresses $A$ as $A = LU$ where $L$ is lower triangular with 1s on the diagonal and $U$ is upper triangular.** 3. **Assume:** $$L = \begin{pmatrix} 1 & 0 \\ l_{21} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{pmatrix}$$ 4. **From $A = LU$, multiply:** $$\begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ l_{21} & 1 \end{pmatrix} \begin{pmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{pmatrix} = \begin{pmatrix} u_{11} & u_{12} \\ l_{21} u_{11} & l_{21} u_{12} + u_{22} \end{pmatrix}$$ 5. **Equate entries:** - $u_{11} = 2$ - $u_{12} = 0$ - $l_{21} u_{11} = 1 \implies l_{21} = \frac{1}{2}$ - $l_{21} u_{12} + u_{22} = 2 \implies 0 + u_{22} = 2 \implies u_{22} = 2$ 6. **Final LU decomposition:** $$L = \begin{pmatrix} 1 & 0 \\ \frac{1}{2} & 1 \end{pmatrix}, \quad U = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$$