Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 14E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 14E
Chapter:
Problem:
Let...Is u in the subset of R3 spanned by the columns of A? Why or why not?
Step-by-Step Solution
Given information
We are given following vector and Matrix:
\[\begin{array}{l}{\bf{u}} = \left[ {\begin{array}{*{20}{r}}2\\{ - 3}\\2\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{r}}5&8&7\\0&1&{ - 1}\\1&3&0\end{array}} \right]\end{array}\]We have to find whether u is subset of $R^3$ spanned by the columns of A or not?
Step 1: Augmented Matrix
\[M = [A\,\,\,\,\,{\bf{u}}] = \left[ {\begin{array}{*{20}{c}}5&8&7&:&2\\0&1&{ - 1}&:&{ - 3}\\1&3&0&:&2\end{array}} \right]\]
Step 2: Row reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}5&8&7&:&2\\0&1&{ - 1}&:&{ - 3}\\1&3&0&:&2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\5&8&7&:&2\end{array}} \right]\,\,::\,\,\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\0&{ - 7}&7&:&{ - 8}\end{array}} \right]\,\,::\,\,\left\{ {{R_3} = {R_3} - 5{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\0&0&0&:&{ - 29}\end{array}} \right]\,\,::\,\,\left\{ {{R_3} = {R_3} + 7{R_2}} \right\}\end{array}\]We can see that the last row can not be true ($0 \ne - 29$), hence the system has no solution. Therefore, $u$ is not in the subset of $R^3$ spanned by the columns of A.
We are given following vector and Matrix:
\[\begin{array}{l}{\bf{u}} = \left[ {\begin{array}{*{20}{r}}2\\{ - 3}\\2\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{r}}5&8&7\\0&1&{ - 1}\\1&3&0\end{array}} \right]\end{array}\]We have to find whether u is subset of $R^3$ spanned by the columns of A or not?
Step 1: Augmented Matrix
\[M = [A\,\,\,\,\,{\bf{u}}] = \left[ {\begin{array}{*{20}{c}}5&8&7&:&2\\0&1&{ - 1}&:&{ - 3}\\1&3&0&:&2\end{array}} \right]\]
Step 2: Row reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}5&8&7&:&2\\0&1&{ - 1}&:&{ - 3}\\1&3&0&:&2\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\5&8&7&:&2\end{array}} \right]\,\,::\,\,\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\0&{ - 7}&7&:&{ - 8}\end{array}} \right]\,\,::\,\,\left\{ {{R_3} = {R_3} - 5{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&0&:&2\\0&1&{ - 1}&:&{ - 3}\\0&0&0&:&{ - 29}\end{array}} \right]\,\,::\,\,\left\{ {{R_3} = {R_3} + 7{R_2}} \right\}\end{array}\]We can see that the last row can not be true ($0 \ne - 29$), hence the system has no solution. Therefore, $u$ is not in the subset of $R^3$ spanned by the columns of A.