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 21E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 21E
Chapter:
Problem:
With A as in Exercise 17, find a nonzero vector in Nul A and a nonzero vector in Col A. Exercise 17: For the matrices in Exercises 17–20, (a) find ksuch that Nul A is
Step-by-Step Solution
Given Information
We are give with following matrix:\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 6}\\{ - 1}&3\\{ - 4}&{12}\\3&{ - 9}\end{array}} \right]\]We have to find a nonzero vector in Nul (A) and a nonzero matrix in Col (A)
Step-1: Augmented matrix for Nul(A)
For Null Space we have to solve Ax=0. The system of equations are: \[\left[ {\begin{array}{*{20}{c}}2&{ - 6}\\{ - 1}&3\\{ - 4}&{12}\\3&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\\0\\0\end{array}} \right]\]The augmented matrix is given by:\[M = \left[ {\begin{array}{*{20}{c}}2&{ - 6}&0\\{ - 1}&3&0\\{ - 4}&{12}&0\\3&{ - 9}&0\end{array}} \right]\]
Step-2: Row-Reduced Augmented matrix
The row-reduced form is given by:\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&{ - 6}&0\\{ - 1}&3&0\\{ - 4}&{12}&0\\3&{ - 9}&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{12}&0\\{ - 1}&3&0\\2&{ - 6}&0\\3&{ - 9}&0\end{array}} \right]::\,\,\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{12}&0\\0&0&0\\0&0&0\\0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} - \dfrac{{{R_1}}}{4}\\{R_3} = {R_3} + \dfrac{{{R_1}}}{2}\\{R_4} = {R_4} + \dfrac{{3{R_1}}}{4}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}{ - 1}&3&0\\0&0&0\\0&0&0\\0&0&0\end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{4}} \right\}\end{array}\]
Step-3: The Nul space
\[ - {x_1} + 3{x_2} = 0\]Since there are two variables and 1 equation, we have to consider $x_2$ as a free variable.\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3t}\\t\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]t\]So, Null Space of A is:
\[{\rm{Nul}}(A) = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\]
Step-4: The Col space
From the Row-Reduced form, we could see that there is only one pivot element that corresponds to column-1. So the Column space of A is given by the first column of matrix A. \[{\rm{Col}}\left( A \right)\, = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\{ - 4}\\3\end{array}} \right]\]
We are give with following matrix:\[A = \left[ {\begin{array}{*{20}{c}}2&{ - 6}\\{ - 1}&3\\{ - 4}&{12}\\3&{ - 9}\end{array}} \right]\]We have to find a nonzero vector in Nul (A) and a nonzero matrix in Col (A)
Step-1: Augmented matrix for Nul(A)
For Null Space we have to solve Ax=0. The system of equations are: \[\left[ {\begin{array}{*{20}{c}}2&{ - 6}\\{ - 1}&3\\{ - 4}&{12}\\3&{ - 9}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\\0\\0\end{array}} \right]\]The augmented matrix is given by:\[M = \left[ {\begin{array}{*{20}{c}}2&{ - 6}&0\\{ - 1}&3&0\\{ - 4}&{12}&0\\3&{ - 9}&0\end{array}} \right]\]
Step-2: Row-Reduced Augmented matrix
The row-reduced form is given by:\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&{ - 6}&0\\{ - 1}&3&0\\{ - 4}&{12}&0\\3&{ - 9}&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{12}&0\\{ - 1}&3&0\\2&{ - 6}&0\\3&{ - 9}&0\end{array}} \right]::\,\,\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}{ - 4}&{12}&0\\0&0&0\\0&0&0\\0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} - \dfrac{{{R_1}}}{4}\\{R_3} = {R_3} + \dfrac{{{R_1}}}{2}\\{R_4} = {R_4} + \dfrac{{3{R_1}}}{4}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}{ - 1}&3&0\\0&0&0\\0&0&0\\0&0&0\end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{4}} \right\}\end{array}\]
Step-3: The Nul space
\[ - {x_1} + 3{x_2} = 0\]Since there are two variables and 1 equation, we have to consider $x_2$ as a free variable.\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3t}\\t\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]t\]So, Null Space of A is:
\[{\rm{Nul}}(A) = \left[ {\begin{array}{*{20}{c}}3\\1\end{array}} \right]\]
Step-4: The Col space
From the Row-Reduced form, we could see that there is only one pivot element that corresponds to column-1. So the Column space of A is given by the first column of matrix A. \[{\rm{Col}}\left( A \right)\, = \left[ {\begin{array}{*{20}{c}}2\\{ - 1}\\{ - 4}\\3\end{array}} \right]\]