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 131: 5

Answer

$y=\begin{bmatrix} 1\\ 5\\ 1\\ -3\\ \end{bmatrix}$, $x=\begin{bmatrix} -2\\ -1\\ 2\\ -3\\ \end{bmatrix}$

Work Step by Step

$Ax=b$ $LUx=b$ First, row reduce the augmented matrix $[L\ b]$ to calculate $Ux$. $\begin{bmatrix} 1&0&0&0&1\\ 2&1&0&0&7\\ -1&0&1&0&0\\ -4&3&-5&1&3\\ \end{bmatrix}$ $y=Ux=\begin{bmatrix} 1\\ 5\\ 1\\ -3\\ \end{bmatrix}$ Next, row reduce augmented matrix $[Uy]$ to calculate $x$. $\begin{bmatrix} 1&-2&-4&-3&1\\ 0&-3&1&0&5\\ 0&0&2&1&1\\ 0&0&0&1&-3\\ \end{bmatrix}$ $x=\begin{bmatrix} -2\\ -1\\ 2\\ -3\\ \end{bmatrix}$
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