Linear Algebra and Its Applications (5th Edition)

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

Chapter 1 - Linear Equations in Linear Algebra - 1.4 Exercises - Page 42: 40

Answer

$\sim \left[\begin{array}{ c c c c c }5&11&-6&\ -7&12\\0&62/5&-62/5&-19/5&39/5\\0&0&0&16/31&-537/31\\0&0&0&\ \ \ 0&1367/62\end{array}\right]$ The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4

Work Step by Step

$\left[\begin{array}{ c c c c c }5&11&-6&-7&12\\-7&-3&-4&6&-9\\11&5&6&-9&-3\\-3&4&-7&2&7\end{array}\right]\sim \left[\begin{array}{ c c c c c }5&11&-6&-7&12\\0&62/5&-62/5&-19/5&39/5\\0&-96/5&96/5&32/5&-147/5\\0&53/5&-53/5&-11/5&71/5\end{array}\right]$ $\sim \left[\begin{array}{ c c c c c }5&11&-6&-7&12\\0&62/5&-62/5&-19/5&39/5\\0&0&0&16/31&-537/31\\0&0&0&65/62&467/62\end{array}\right]\sim \left[\begin{array}{ c c c c c }5&11&-6&\ -7&12\\0&62/5&-62/5&-19/5&39/5\\0&0&0&16/31&-537/31\\0&0&0&\ \ \ 0&1367/62\end{array}\right]$ The original matrix has a pivot in every row, so its columns span $R^4$, by Theorem 4
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