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 5E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Find an explicit description of Nul A by listing vectors that span the null space...
Step-by-Step Solution
Step 1
Given matrix\[A = \left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&4&0\\0&0&1&{ - 9}&0\\0&0&0&0&1\end{array}} \right]\]We have to find an explicit description of ${\rm{Nul}}(A)$ by listingvectors that span the null space
Step 2: The matrix form
The basis for the ${\rm{Nul}}(A)$ is the solution space of the system $A{\bf{x}} = b$. Write the matrix form of system of equations:\[\begin{array}{l}A{\bf{x}} = b\\\left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&4&0\\0&0&1&{ - 9}&0\\0&0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\]
Step 3: Equation form
The equations for above matrix form are: \[\left\{ \begin{array}{l}{x_1} - 2{x_2} + 4{x_4} = 0,\\{x_3} - 9{x_4} = 0,\\{x_5} = 0\end{array} \right.\]
Step 4: parametric Form
\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{2{x_2} - 4{x_4}}\\{{x_2}}\\{9{x_4}}\\{{x_4}}\\0\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}}2\\1\\0\\0\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\9\\1\\0\end{array}} \right]\end{array}\]
ANSWER
Hence the vectors that span ${\rm{Nul}}(A)$ are: \[\left\{ {\left[ {\begin{array}{*{20}{l}}2\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\9\\1\\0\end{array}} \right]} \right\}\]
Given matrix\[A = \left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&4&0\\0&0&1&{ - 9}&0\\0&0&0&0&1\end{array}} \right]\]We have to find an explicit description of ${\rm{Nul}}(A)$ by listingvectors that span the null space
Step 2: The matrix form
The basis for the ${\rm{Nul}}(A)$ is the solution space of the system $A{\bf{x}} = b$. Write the matrix form of system of equations:\[\begin{array}{l}A{\bf{x}} = b\\\left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&4&0\\0&0&1&{ - 9}&0\\0&0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\]
Step 3: Equation form
The equations for above matrix form are: \[\left\{ \begin{array}{l}{x_1} - 2{x_2} + 4{x_4} = 0,\\{x_3} - 9{x_4} = 0,\\{x_5} = 0\end{array} \right.\]
Step 4: parametric Form
\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{2{x_2} - 4{x_4}}\\{{x_2}}\\{9{x_4}}\\{{x_4}}\\0\end{array}} \right]\\ = {x_2}\left[ {\begin{array}{*{20}{l}}2\\1\\0\\0\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\9\\1\\0\end{array}} \right]\end{array}\]
ANSWER
Hence the vectors that span ${\rm{Nul}}(A)$ are: \[\left\{ {\left[ {\begin{array}{*{20}{l}}2\\1\\0\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 4}\\0\\9\\1\\0\end{array}} \right]} \right\}\]