Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 23E from Chapter 2.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 23E
Chapter:
Problem:
Exercises 23–26 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 a Matrix and its echelon form\[A = \left[ {\begin{array}{*{20}{l}}4&5&9&{ - 2}\\6&5&1&{12}\\3&4&8&{ - 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\]We have to find the basis for Col A and a basis for Nul A
Step 1: Basis for Col A
Basis for Col A are only non-zero rows of the echelon form of the matrix. we can see from the echelon form that, first two columns have pivot elements.
So basis of Col A is:Therefore,
\[{\rm{Col}}\,A\, = \left\{ {\left[ {\begin{array}{*{20}{c}}4\\6\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\5\\4\end{array}} \right]} \right\}\]
Step 2: Basis for Nul A
the basis for Nul A is given by the solution vector of $Ax=0$
Matrix form of Ax=0\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{l}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\]System of Equations\[\begin{array}{l}{x_2} + 5{x_3} - 6{x_4} = 0\\{x_1} + 2\left( { - 5{x_3} + 6{x_4}} \right) + 6{x_3} - 5{x_4} = 0\end{array}\]
Step 3: The Solution
Considering, $x_3$ and $x_4$ as free variables,\[\begin{array}{l}{x_2} = - 5{x_3} + 6{x_4}\\{x_1} = 4{x_3} - 7{x_4}\end{array}\]The Solution Set: \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{x_3} - 7{x_4}}\\{ - 5{x_3} + 6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}}{4{x_3}}\\{ - 5{x_3}}\\{{x_3}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7{x_4}}\\{6{x_4}}\\0\\{{x_4}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{r}}4\\{ - 5}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{r}}{ - 7}\\6\\0\\1\end{array}} \right]\end{array}\] Therefore,
\[{\rm{Nul}}\,A = \left\{ {\left[ {\begin{array}{*{20}{r}}4\\{ - 5}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{r}}{ - 7}\\6\\0\\1\end{array}} \right]} \right\}\]
We are given with a Matrix and its echelon form\[A = \left[ {\begin{array}{*{20}{l}}4&5&9&{ - 2}\\6&5&1&{12}\\3&4&8&{ - 3}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\]We have to find the basis for Col A and a basis for Nul A
Step 1: Basis for Col A
Basis for Col A are only non-zero rows of the echelon form of the matrix. we can see from the echelon form that, first two columns have pivot elements.
So basis of Col A is:Therefore,
\[{\rm{Col}}\,A\, = \left\{ {\left[ {\begin{array}{*{20}{c}}4\\6\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}5\\5\\4\end{array}} \right]} \right\}\]
Step 2: Basis for Nul A
the basis for Nul A is given by the solution vector of $Ax=0$
Matrix form of Ax=0\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{l}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\end{array}\]System of Equations\[\begin{array}{l}{x_2} + 5{x_3} - 6{x_4} = 0\\{x_1} + 2\left( { - 5{x_3} + 6{x_4}} \right) + 6{x_3} - 5{x_4} = 0\end{array}\]
Step 3: The Solution
Considering, $x_3$ and $x_4$ as free variables,\[\begin{array}{l}{x_2} = - 5{x_3} + 6{x_4}\\{x_1} = 4{x_3} - 7{x_4}\end{array}\]The Solution Set: \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{x_3} - 7{x_4}}\\{ - 5{x_3} + 6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}}{4{x_3}}\\{ - 5{x_3}}\\{{x_3}}\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7{x_4}}\\{6{x_4}}\\0\\{{x_4}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{r}}4\\{ - 5}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{r}}{ - 7}\\6\\0\\1\end{array}} \right]\end{array}\] Therefore,
\[{\rm{Nul}}\,A = \left\{ {\left[ {\begin{array}{*{20}{r}}4\\{ - 5}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{r}}{ - 7}\\6\\0\\1\end{array}} \right]} \right\}\]