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 3E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 3E
Chapter:
Problem:
0
Step-by-Step Solution
Given Information
We are given with following matrix: \[A=\left[\begin{array}{llll}1 & 3 & 5 & 0 \\0 & 1 & 4 & -2\end{array}\right]\]We have to find an explicit description of Nul A by listing vectors that span the null space. The null space is the set of all homogeneous solutions of $Ax=0$.
Step-1: General solution of $Ax=0$
Matrix form of equations:\[\begin{array}{*{20}{l}}{A{\bf{x}} = 0}\\{\left[ {\begin{array}{*{20}{c}}1&3&5&0\\0&1&4&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\end{array}} \right]}\end{array}\]System of equations: \[\begin{array}{*{20}{l}}\begin{array}{l}{x_1} - 7{x_3} + 6{x_4} = 0\\\end{array}\\{{x_2} + 4{x_3} - 2{x_4} = 0 = 0}\end{array}\]Consider $x_3$ and $x_4$ as free, \[\begin{array}{*{20}{l}}{{x_1} = 7{x_3} - 6{x_4}}\\{{x_2} = - 4{x_3} + 2{x_4}}\\{{x_3} = s}\\{{x_4} = t}\end{array}\]So, solution in parametric form is:\[\begin{array}{*{20}{l}}\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{7{x_3} - 6{x_4}}\\{ - 4{x_3} + 2{x_4}}\\s\\t\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{7s - 6t}\\{ - 4s + 2t}\\s\\t\end{array}} \right]\end{array}\\{ = \left[ {\begin{array}{*{20}{c}}7\\{ - 4}\\1\\0\end{array}} \right]s + \left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\0\\1\end{array}} \right]t}\end{array}\]Therefore, the Nul space is:
\[Nul{\mkern 1mu} A = \left\{ {\left[ {\begin{array}{*{20}{c}}7\\{ - 4}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\0\\1\end{array}} \right]} \right\}\]
We are given with following matrix: \[A=\left[\begin{array}{llll}1 & 3 & 5 & 0 \\0 & 1 & 4 & -2\end{array}\right]\]We have to find an explicit description of Nul A by listing vectors that span the null space. The null space is the set of all homogeneous solutions of $Ax=0$.
Step-1: General solution of $Ax=0$
Matrix form of equations:\[\begin{array}{*{20}{l}}{A{\bf{x}} = 0}\\{\left[ {\begin{array}{*{20}{c}}1&3&5&0\\0&1&4&{ - 2}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\end{array}} \right]}\end{array}\]System of equations: \[\begin{array}{*{20}{l}}\begin{array}{l}{x_1} - 7{x_3} + 6{x_4} = 0\\\end{array}\\{{x_2} + 4{x_3} - 2{x_4} = 0 = 0}\end{array}\]Consider $x_3$ and $x_4$ as free, \[\begin{array}{*{20}{l}}{{x_1} = 7{x_3} - 6{x_4}}\\{{x_2} = - 4{x_3} + 2{x_4}}\\{{x_3} = s}\\{{x_4} = t}\end{array}\]So, solution in parametric form is:\[\begin{array}{*{20}{l}}\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{7{x_3} - 6{x_4}}\\{ - 4{x_3} + 2{x_4}}\\s\\t\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{7s - 6t}\\{ - 4s + 2t}\\s\\t\end{array}} \right]\end{array}\\{ = \left[ {\begin{array}{*{20}{c}}7\\{ - 4}\\1\\0\end{array}} \right]s + \left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\0\\1\end{array}} \right]t}\end{array}\]Therefore, the Nul space is:
\[Nul{\mkern 1mu} A = \left\{ {\left[ {\begin{array}{*{20}{c}}7\\{ - 4}\\1\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 6}\\2\\0\\1\end{array}} \right]} \right\}\]