# Recitation 6B

Problem 1: Solve the equation A\mathbf{x}=\mathbf{b} by using the LU factorization for A=\begin{bmatrix}2&-4&2\\-4&5&2\\6&-9&1\end{bmatrix}=\begin{bmatrix}1&0&0\\-2&1&0\\3&-1&1\end{bmatrix}\begin{bmatrix}2&-4&2\\0&-3&6\\0&0&1\end{bmatrix} and \mathbf{b}=\begin{bmatrix}6\\0\\6\end{bmatrix}.

Solution: Let L=\begin{bmatrix}1&0&0\\-2&1&0\\3&-1&1\end{bmatrix}, U=\begin{bmatrix}2&-4&2\\0&-3&6\\0&0&1\end{bmatrix}. First we would like to solve L\mathbf{y}=\mathbf{b}. By forward substitution, we get y_1 = 6, y_2=0+2y_1=12, y_3=6-3y_1+y_2=0. Finally we would like to solve U\mathbf{x}=\mathbf{y}. By backward substitution, we get x_3=0, x_2=(12-6x_3)/(-3)=-4, x_1=(6+4x_2-2x_3)/2=-5.

Problem 2: Find an LU factorization of A=\begin{bmatrix}2&-4&-2&3\\6&-9&-5&8\\2&-7&-3&9\\4&-2&-2&-1\\-6&3&3&4\end{bmatrix}.

Solution: Start row-reducing the matrix A: \begin{bmatrix}\mathbf{2}&-4&-2&3\\\mathbf{6}&-9&-5&8\\\mathbf{2}&-7&-3&9\\\mathbf{4}&-2&-2&-1\\\mathbf{-6}&3&3&4\end{bmatrix}\sim \begin{bmatrix}2&-4&-2&3\\0&\mathbf{3}&1&-1\\0&\mathbf{-3}&-1&6\\0&\mathbf{6}&2&-7\\0&\mathbf{-9}&-3&13\end{bmatrix}\sim \begin{bmatrix}2&-4&-2&3\\0&3&1&-1\\0&0&0&\mathbf{5}\\0&0&0&\mathbf{-5}\\0&0&0&\mathbf{10}\end{bmatrix}\sim \begin{bmatrix}2&-4&-2&3\\0&3&1&-1\\0&0&0&5\\0&0&0&0\\0&0&0&0\end{bmatrix}=U Pick out the entries in boldface, normalized and assemble them to make L=\begin{bmatrix}1 & 0 & 0 & 0 & 0\\ 3 & 1 & 0 & 0 & 0\\ 1 & -1 & 1 & 0 & 0\\ 2 & 2 & -1 & 1 & 0 \\ -3 & -3 & 2 & 0 & 1\end{bmatrix}.

Problem 3: Consider the production model \mathbf{x}=C\mathbf{x}+\mathbf{d} for an economy with two sectors, where C=\begin{bmatrix}.0 & .5\\ .6 & .2\end{bmatrix}, \mathbf{b}=\begin{bmatrix}50\\30\end{bmatrix}. Use an inverse matrix to determine the production level necessary to satisfy the final demand.

Solution(sketch): According to the equation, we obtain (I-C)\mathbf{x}=\mathbf{d}. Then use Gaussian elimination to solve it.