Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises: 3

Answer

$\begin{bmatrix} 1&0&-1&-2\\ 0&1&2&3\\ 0&0&0&0 \end{bmatrix}$ The pivots are the the entries in positions $(1, 1)$ and $(2, 2)$ in both the original and the final matrices, where coordinates are in $(row, column)$ form. The pivot columns are therefore columns one and two.

Work Step by Step

(1) Begin with the original matrix. $ \begin{bmatrix} 1&2&3&4\\ 4&5&6&7\\ 6&7&8&9 \end{bmatrix} $ (2) Add $-4$ times row one to row two. Add $-6$ times row one to row three. $ \begin{bmatrix} 1&2&3&4\\ 0&-3&-6&-9\\ 0&-5&-10&-15 \end{bmatrix}$ (3) Multiply row two by $-\frac{1}{3}$. Multiply row three by $-\frac{1}{5}$. $ \begin{bmatrix} 1&2&3&4\\ 0&1&2&3\\ 0&1&2&3 \end{bmatrix} $ (4) Add $-1$ times row two to row three. Add $-2$ times row two to row one. $ \begin{bmatrix} 1&0&-1&-2\\ 0&1&2&3\\ 0&0&0&0 \end{bmatrix} $
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