Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 2 - Matrix Algebra - 2.5 Exercises - Page 132: 11

Answer

The LU factorization: $\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\2&1&0\\-\frac{1}{3}&1&1\end{bmatrix}\begin{bmatrix}3&-6&3\\0&5&-4\\0&0&5\end{bmatrix}$

Work Step by Step

To find LU factorization of the matrix $\mathbf{A}=\begin{bmatrix}3&-6&3\\6&-7&2\\-1&7&0\end{bmatrix}$ Let's Reduce the rows of the matrix; Let $R_{2}=R_{2}-2R_{1}$ =$\begin{bmatrix}3&-6&3\\0&5&-4\\-1&7&0\end{bmatrix}$ Let $R_{3}=R_{3}+\frac{R_{1}}{3}$ =$\begin{bmatrix}3&-6&3\\0&5&-4\\0&5&1\end{bmatrix}$ Let $R_{3}=R_{3}-R_{2}$ =$\begin{bmatrix}3&-6&3\\0&5&-4\\0&0&5\end{bmatrix}$ Hence; $\mathbf{U}=\begin{bmatrix}3&-6&3\\0&5&-4\\0&0&5\end{bmatrix}$ To find the matrix L we divide each column by its pivot entry; $\mathbf{L}=\begin{bmatrix}\frac{3}{3}&0&0\\\frac{6}{3}&\frac{5}{5}&0\\-\frac{1}{3}&\frac{5}{5}&\frac{5}{5}\end{bmatrix}=\begin{bmatrix}1&0&0\\2&1&0\\-\frac{1}{3}&1&1\end{bmatrix}$ Hence, $\mathbf{A}=\mathbf{LU}=\begin{bmatrix}1&0&0\\2&1&0\\-\frac{1}{3}&1&1\end{bmatrix}\begin{bmatrix}3&-6&3\\0&5&-4\\0&0&5\end{bmatrix}$
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