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 10E from Chapter 1.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 10E
Chapter:
Problem:
Find all...that are mapped into the zero vector by the transformation...for the given matrix A...
Step-by-Step Solution
Given information
We are given with following Matrix: \[A = \left[ {\begin{array}{*{20}{r}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\]We have to find x in the mapping \[T:{R^4} \to {R^4},\,\,T\left( x \right) = Ax\]
Step 1: Matrix form of Equations
\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{r}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\\0\\0\end{array}} \right]\end{array}\]
Step 2: Augmented Matrix
\[M = \left[ {A\,\,\,0} \right] = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\]
Step 3: Row Reduced Augmented Matrix
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\0&9&{18}&9&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} \to {R_2} - {R_1}\\{R_4} \to {R_4} + 2{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&9&{18}&9&0\end{array}} \right]::\,\,\left\{ {{R_2} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&0&0&{ - 18}&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_3} \to {R_3} + 3{R_2}\\{R_4} \to {R_4} - 9{R_2}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{l}}1&3&9&0&0\\0&1&2&0&0\\0&0&0&3&0\\0&0&0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} - \dfrac{{2{R_2}}}{3}\\{R_2} \to {R_2} - {R_3}\\{R_4} \to {R_4} + 6{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_3} \to \dfrac{{{R_3}}}{3}\\{R_1} \to {R_1} - 3{R_2}\end{array} \right\}\end{array}\]So,
\[M = \left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\]
Step 4: General system of Equations
Since there are three pivot elements and 4 variables, we will consider on of them as free variable. The equations are\[{x_1} + 3{x_3} = 0,{x_2} + 2{x_3} = 0,{x_4} = 0\]Considering $x_3$ as free:\[\begin{array}{l}{x_1} = - 3{x_3},\\{x_2} = - 2{x_3},\\{x_3} = {x_3}\\{x_4} = 0,{x_3}\end{array}\]\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3{x_3}}\\{ - 2{x_3}}\\{{x_3}}\\0\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\end{array}\]So,
\[{\bf{x}} = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\]
We are given with following Matrix: \[A = \left[ {\begin{array}{*{20}{r}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\]We have to find x in the mapping \[T:{R^4} \to {R^4},\,\,T\left( x \right) = Ax\]
Step 1: Matrix form of Equations
\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{r}}1&3&9&2\\1&0&3&{ - 4}\\0&1&2&3\\{ - 2}&3&0&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}0\\0\\0\\0\end{array}} \right]\end{array}\]
Step 2: Augmented Matrix
\[M = \left[ {A\,\,\,0} \right] = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\]
Step 3: Row Reduced Augmented Matrix
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\1&0&3&{ - 4}&0\\0&1&2&3&0\\{ - 2}&3&0&5&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&1&2&3&0\\0&9&{18}&9&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} \to {R_2} - {R_1}\\{R_4} \to {R_4} + 2{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&{ - 3}&{ - 6}&{ - 6}&0\\0&9&{18}&9&0\end{array}} \right]::\,\,\left\{ {{R_2} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&9&2&0\\0&1&2&3&0\\0&0&0&3&0\\0&0&0&{ - 18}&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_3} \to {R_3} + 3{R_2}\\{R_4} \to {R_4} - 9{R_2}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{l}}1&3&9&0&0\\0&1&2&0&0\\0&0&0&3&0\\0&0&0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} - \dfrac{{2{R_2}}}{3}\\{R_2} \to {R_2} - {R_3}\\{R_4} \to {R_4} + 6{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_3} \to \dfrac{{{R_3}}}{3}\\{R_1} \to {R_1} - 3{R_2}\end{array} \right\}\end{array}\]So,
\[M = \left[ {\begin{array}{*{20}{c}}1&0&3&0&0\\0&1&2&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\]
Step 4: General system of Equations
Since there are three pivot elements and 4 variables, we will consider on of them as free variable. The equations are\[{x_1} + 3{x_3} = 0,{x_2} + 2{x_3} = 0,{x_4} = 0\]Considering $x_3$ as free:\[\begin{array}{l}{x_1} = - 3{x_3},\\{x_2} = - 2{x_3},\\{x_3} = {x_3}\\{x_4} = 0,{x_3}\end{array}\]\[\begin{array}{l}{\bf{x}} = \left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3{x_3}}\\{ - 2{x_3}}\\{{x_3}}\\0\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\end{array}\]So,
\[{\bf{x}} = {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\\0\end{array}} \right]\]