Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 893: 20

Answer

$\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 5 & -10 & -5 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right],\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 1 & -2 & -1 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]$

Work Step by Step

A matrix in which all the diagonal elements are $1$ and the all the elements lower than the principal diagonal are $0$ is called a matrix in row-echelon form. The given matrix can be converted into the row-echelon form by performing some matrix operations. This is done as follows: Perform the operation $-2{{R}_{1}}+{{R}_{2}}$ to get $\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ -2+2 & 4+1 & -6-4 & -8+3 \\ -3 & 4 & -1 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 5 & -10 & -5 \\ -3 & 4 & -1 & -2 \\ \end{matrix} \right]$ Now perform the operation $3{{R}_{1}}+{{R}_{3}}$ to get $\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 5 & -10 & -5 \\ 3-3 & -6+4 & 9-1 & 12-2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 5 & -10 & -5 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]$ The matrix obtained above is similar to the matrix provided with missing elements. Thus, the first solution matrix is given by, $\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 5 & -10 & -5 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]$ Now perform the operation $\frac{1}{5}{{R}_{2}}$ to get $\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & \frac{1}{5}\times 5 & \frac{1}{5}\times (-10) & \frac{1}{5}\times (-5) \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 1 & -2 & -1 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]$ Thus, the second solution matrix is given by, $\left[ \begin{matrix} 1 & -2 & 3 & 4 \\ 0 & 1 & -2 & -1 \\ 0 & -2 & 8 & 10 \\ \end{matrix} \right]$
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