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 24E from Chapter 2.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 24E
Chapter:
Problem:
Exercises display a matrix A and an echelon form of A. Find a basis for Col A and a basis for Nul A.
Step-by-Step Solution
Given Information
We are given with following matrix: \[A = \left[ {\begin{array}{*{20}{c}} { - 3}&9&{ - 2}&{ - 7}\\ 2&{ - 6}&4&8\\ 3&{ - 9}&{ - 2}&2 \end{array}} \right]\] We have to find the basis for its row and column space. We are also given with its equivalent echelon form. \[A = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&6&9\\ 0&0&4&5\\ 0&0&0&0 \end{array}} \right]\]
Step-1: Basis for Col A
As we can see that there are two columns (first and third) with pivot elements. Hence, the column space is given by: \[{\rm{Col}}\,\,{\rm{A}}\,\,{\rm{ = }}\left\{ {\left[ {\begin{array}{*{20}{c}} { - 3}\\ 2\\ 3 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} { - 2}\\ 4\\ { - 2} \end{array}} \right]} \right\}\]
Step-2: Basis for Nul A
Solve the system of equations $Ax=0$ \[\begin{array}{l} Ax = 0\\ \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&6&9\\ 0&0&4&5\\ 0&0&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The System of equations are: \[ \begin{array} { r } { x _ { 1 } - 3 x _ { 2 } + 6 x _ { 3 } + 9 x _ { 4 } = 0 } \\ { x _ { 3 } + \dfrac { 5 } { 4 } x _ { 4 } = 0 } \end{array} \] Substitute $x_3$ from second equation to first. \[\begin{array}{l} {x_1} = 3{x_2} - 6{x_3} - 9{x_4}\\ {x_1} = 3{x_2} - 6\left( { - \dfrac{5}{4}{x_4}} \right) - 9{x_4}\\ {x_1} = 3{x_2} - \dfrac{3}{2}{x_4} \end{array}\] Upon solving them \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3{x_2} - \dfrac{3}{2}{x_4}}\\ {{x_2}}\\ { - \dfrac{5}{4}{x_4}}\\ {{x_4}} \end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}} {\dfrac{{ - 3}}{2}}\\ 0\\ { - \dfrac{5}{4}}\\ 1 \end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right] - 4{x_4}\left[ {\begin{array}{*{20}{c}} 6\\ 0\\ 5\\ { - 4} \end{array}} \right] \end{array}\] therefore, the basis for Null space of A is:
\[{\rm{Nul}}\,\,A = \left\{ {\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}} 6\\ 0\\ 5\\ { - 4} \end{array}} \right]} \right\}\]
We are given with following matrix: \[A = \left[ {\begin{array}{*{20}{c}} { - 3}&9&{ - 2}&{ - 7}\\ 2&{ - 6}&4&8\\ 3&{ - 9}&{ - 2}&2 \end{array}} \right]\] We have to find the basis for its row and column space. We are also given with its equivalent echelon form. \[A = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&6&9\\ 0&0&4&5\\ 0&0&0&0 \end{array}} \right]\]
Step-1: Basis for Col A
As we can see that there are two columns (first and third) with pivot elements. Hence, the column space is given by: \[{\rm{Col}}\,\,{\rm{A}}\,\,{\rm{ = }}\left\{ {\left[ {\begin{array}{*{20}{c}} { - 3}\\ 2\\ 3 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} { - 2}\\ 4\\ { - 2} \end{array}} \right]} \right\}\]
Step-2: Basis for Nul A
Solve the system of equations $Ax=0$ \[\begin{array}{l} Ax = 0\\ \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&6&9\\ 0&0&4&5\\ 0&0&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 0 \end{array}} \right] \end{array}\] The System of equations are: \[ \begin{array} { r } { x _ { 1 } - 3 x _ { 2 } + 6 x _ { 3 } + 9 x _ { 4 } = 0 } \\ { x _ { 3 } + \dfrac { 5 } { 4 } x _ { 4 } = 0 } \end{array} \] Substitute $x_3$ from second equation to first. \[\begin{array}{l} {x_1} = 3{x_2} - 6{x_3} - 9{x_4}\\ {x_1} = 3{x_2} - 6\left( { - \dfrac{5}{4}{x_4}} \right) - 9{x_4}\\ {x_1} = 3{x_2} - \dfrac{3}{2}{x_4} \end{array}\] Upon solving them \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3{x_2} - \dfrac{3}{2}{x_4}}\\ {{x_2}}\\ { - \dfrac{5}{4}{x_4}}\\ {{x_4}} \end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}} {\dfrac{{ - 3}}{2}}\\ 0\\ { - \dfrac{5}{4}}\\ 1 \end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right] - 4{x_4}\left[ {\begin{array}{*{20}{c}} 6\\ 0\\ 5\\ { - 4} \end{array}} \right] \end{array}\] therefore, the basis for Null space of A is:
\[{\rm{Nul}}\,\,A = \left\{ {\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}} 6\\ 0\\ 5\\ { - 4} \end{array}} \right]} \right\}\]