1. **Problem 1:** Find the standard matrix $A$ representing the linear operator $L: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $L\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}x + 2y \\ x + y\end{pmatrix}$. 2. The standard matrix $A$ is formed by applying $L$ to the standard basis vectors $e_1 = \begin{pmatrix}1 \\ 0\end{pmatrix}$ and $e_2 = \begin{pmatrix}0 \\ 1\end{pmatrix}$ and placing the results as columns: $$A = \begin{pmatrix}L(e_1) & L(e_2)\end{pmatrix}$$ 3. Calculate $L(e_1)$: $$L\begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 + 2\cdot0 \\ 1 + 0\end{pmatrix} = \begin{pmatrix}1 \\ 1\end{pmatrix}$$ 4. Calculate $L(e_2)$: $$L\begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}0 + 2\cdot1 \\ 0 + 1\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix}$$ 5. Thus, the matrix $A$ is: $$A = \begin{pmatrix}1 & 2 \\ 1 & 1\end{pmatrix}$$ --- 6. **Problem 2:** Determine if matrices $$A = \begin{pmatrix}3 & 2 \\ -1 & 3\end{pmatrix}$$ and $$B = \begin{pmatrix}-1 & -2 \\ 2 & 1\end{pmatrix}$$ commute, i.e., check if $AB = BA$. 7. Compute $AB$: $$AB = \begin{pmatrix}3 & 2 \\ -1 & 3\end{pmatrix} \begin{pmatrix}-1 & -2 \\ 2 & 1\end{pmatrix} = \begin{pmatrix}3\cdot(-1) + 2\cdot2 & 3\cdot(-2) + 2\cdot1 \\ -1\cdot(-1) + 3\cdot2 & -1\cdot(-2) + 3\cdot1\end{pmatrix} = \begin{pmatrix}-3 + 4 & -6 + 2 \\ 1 + 6 & 2 + 3\end{pmatrix} = \begin{pmatrix}1 & -4 \\ 7 & 5\end{pmatrix}$$ 8. Compute $BA$: $$BA = \begin{pmatrix}-1 & -2 \\ 2 & 1\end{pmatrix} \begin{pmatrix}3 & 2 \\ -1 & 3\end{pmatrix} = \begin{pmatrix}-1\cdot3 + (-2)\cdot(-1) & -1\cdot2 + (-2)\cdot3 \\ 2\cdot3 + 1\cdot(-1) & 2\cdot2 + 1\cdot3\end{pmatrix} = \begin{pmatrix}-3 + 2 & -2 - 6 \\ 6 - 1 & 4 + 3\end{pmatrix} = \begin{pmatrix}-1 & -8 \\ 5 & 7\end{pmatrix}$$ 9. Since $AB = \begin{pmatrix}1 & -4 \\ 7 & 5\end{pmatrix} \neq BA = \begin{pmatrix}-1 & -8 \\ 5 & 7\end{pmatrix}$, matrices $A$ and $B$ do not commute. --- 10. **Problem 3:** Find an LU decomposition of matrix $$A = \begin{pmatrix}2 & 0 \\ 1 & 2\end{pmatrix}$$ 11. LU decomposition factors $A$ into $L$ (lower triangular) and $U$ (upper triangular) matrices such that $A = LU$. 12. Let $$L = \begin{pmatrix}1 & 0 \\ l_{21} & 1\end{pmatrix}, \quad U = \begin{pmatrix}u_{11} & u_{12} \\ 0 & u_{22}\end{pmatrix}$$ 13. From $A = LU$, multiply: $$\begin{pmatrix}1 & 0 \\ l_{21} & 1\end{pmatrix} \begin{pmatrix}u_{11} & u_{12} \\ 0 & u_{22}\end{pmatrix} = \begin{pmatrix}u_{11} & u_{12} \\ l_{21}u_{11} & l_{21}u_{12} + u_{22}\end{pmatrix} = \begin{pmatrix}2 & 0 \\ 1 & 2\end{pmatrix}$$ 14. Equate entries: - $u_{11} = 2$ - $u_{12} = 0$ - $l_{21} u_{11} = 1 \implies l_{21} = \frac{1}{2}$ - $l_{21} u_{12} + u_{22} = 2 \implies 0 + u_{22} = 2 \implies u_{22} = 2$ 15. Therefore, $$L = \begin{pmatrix}1 & 0 \\ \frac{1}{2} & 1\end{pmatrix}, \quad U = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$$ **Final answers:** - Problem 1: $A = \begin{pmatrix}1 & 2 \\ 1 & 1\end{pmatrix}$ - Problem 2: $A$ and $B$ do not commute. - Problem 3: $L = \begin{pmatrix}1 & 0 \\ \frac{1}{2} & 1\end{pmatrix}$, $U = \begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}$