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 7E from Chapter 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
In Exercises 7-12, describe all solutions of Ax = 0 in parametric vector form, where A is row equivalent to the given matrix...
Step-by-Step Solution
Step 1
Given Augmented Matrix:
\[M = \left[ {\begin{array}{*{20}{c}}1&3&{ - 3}&7\\0&1&{ - 4}&5\end{array}} \right]\]We have to solve the given augmented matrix and write the solution in parametric vector form.
Step 2: The Augmented Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&{ - 3}&7\\0&1&{ - 4}&5\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&0&9&{ - 8}\\0&1&{ - 4}&5\end{array}} \right]\,\,\,::\,\,\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]
Step 3: System of Equations
We convert the above augmented matrix into system of equations as\[\begin{array}{l}A{\bf{x}} = 0\\\\\left[ {\begin{array}{*{20}{c}}1&0&9&{ - 8}\\0&1&{ - 4}&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\\\\\left\{ {\begin{array}{*{20}{c}}{{x_1} + 9{x_3} - 8{x_4} = 0}\\{{x_2} - 4{x_3} + 5{x_4} = 0}\end{array}} \right\}\end{array}\]
Step 4: Parametric forms
Since there are 4 variables and 2 equations, we have to choose 2 variables as free and parametrize them accordingly. We can solve for $x_1$ and $x_2$ while choosing other as\[{x_3} = r;\,\,\,{x_4} = s; \]
Step 5: General Solution
\[\left\{ \begin{array}{l}{x_1} = - 9{x_3} - 8{x_4} \Rightarrow - 9r - 8s\\{x_2} = 4{x_3} - 5{x_4} \Rightarrow 4r - 5s\end{array} \right.\]
Step 6: Solution in parametric form
Separating the parameters out of the vectors\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 9r - 8s}\\{4r - 5s}\\r\\s\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = r\left[ {\begin{array}{*{20}{c}}{ - 9}\\4\\1\\0\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}8\\{ - 5}\\0\\1\end{array}} \right]\end{array}\]
ANSWER
\[\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = r\left[ {\begin{array}{*{20}{c}}{ - 9}\\4\\1\\0\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}8\\{ - 5}\\0\\1\end{array}} \right]\]
Given Augmented Matrix:
\[M = \left[ {\begin{array}{*{20}{c}}1&3&{ - 3}&7\\0&1&{ - 4}&5\end{array}} \right]\]We have to solve the given augmented matrix and write the solution in parametric vector form.
Step 2: The Augmented Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&{ - 3}&7\\0&1&{ - 4}&5\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&0&9&{ - 8}\\0&1&{ - 4}&5\end{array}} \right]\,\,\,::\,\,\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]
Step 3: System of Equations
We convert the above augmented matrix into system of equations as\[\begin{array}{l}A{\bf{x}} = 0\\\\\left[ {\begin{array}{*{20}{c}}1&0&9&{ - 8}\\0&1&{ - 4}&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\\\\\left\{ {\begin{array}{*{20}{c}}{{x_1} + 9{x_3} - 8{x_4} = 0}\\{{x_2} - 4{x_3} + 5{x_4} = 0}\end{array}} \right\}\end{array}\]
Step 4: Parametric forms
Since there are 4 variables and 2 equations, we have to choose 2 variables as free and parametrize them accordingly. We can solve for $x_1$ and $x_2$ while choosing other as\[{x_3} = r;\,\,\,{x_4} = s; \]
Step 5: General Solution
\[\left\{ \begin{array}{l}{x_1} = - 9{x_3} - 8{x_4} \Rightarrow - 9r - 8s\\{x_2} = 4{x_3} - 5{x_4} \Rightarrow 4r - 5s\end{array} \right.\]
Step 6: Solution in parametric form
Separating the parameters out of the vectors\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 9r - 8s}\\{4r - 5s}\\r\\s\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = r\left[ {\begin{array}{*{20}{c}}{ - 9}\\4\\1\\0\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}8\\{ - 5}\\0\\1\end{array}} \right]\end{array}\]
ANSWER
\[\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = r\left[ {\begin{array}{*{20}{c}}{ - 9}\\4\\1\\0\end{array}} \right] + s\left[ {\begin{array}{*{20}{c}}8\\{ - 5}\\0\\1\end{array}} \right]\]