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 1.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 3E
Chapter:
Problem:
In Exercises with T defined by T(x) = Ax, find a vector × whose image under T is b, and determine whether × is unique...
Step-by-Step Solution
Step 1
Given matrix and Vector:
\[A = \left[ {\begin{array}{*{20}{r}}1&0&{ - 2}\\{ - 2}&1&6\\3&{ - 2}&{ - 5}\end{array}} \right],{\bf{b}} = \left[ {\begin{array}{*{20}{r}}{ - 1}\\7\\{ - 3}\end{array}} \right]\]Given Transformation:\[T:{R^3} \to {R^2}::\,T\left( {\bf{x}} \right) = A{\bf{x}}\]We have to solve the system of equations given by: \[T\left( {\bf{x}} \right) = {\bf{b}}\]
Step 2: The system of Equations
\[\left[ {\begin{array}{*{20}{r}}1&0&{ - 2}\\{ - 2}&1&6\\3&{ - 2}&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ - 1}\\7\\{ - 3}\end{array}} \right]\]
Step 3: The corresponding Augmented matrix form
\[M = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\{ - 2}&1&6&7\\3&{ - 2}&{ - 5}&{ - 3}\end{array}} \right]\]
Step 4: Convert to row reduced echelon form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\{ - 2}&1&6&7\\3&{ - 2}&{ - 5}&{ - 3}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&{ - 2}&1&0\end{array}} \right]::\,{R_3} = {R_3} - 3{R_1};\,\,{R_2} = {R_2} + 2{R_1}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&0&5&{10}\end{array}} \right]::\,{R_3} = {R_3} + 2{R_2}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&0&1&2\end{array}} \right]::\,{R_3} = {R_3}/5\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&0&1\\0&0&1&2\end{array}} \right]::\,{R_3} = {R_2} - 2{R_3}\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&3\\0&1&0&1\\0&0&1&2\end{array}} \right]::\,{R_1} = {R_1} + 2{R_3}\end{array}\]
Step 5: The System of Equations
Since there are three variables and three equations:\[\left\{ \begin{array}{l}{x_1} = 3\\{x_2} = 1\\{x_3} = 2\end{array} \right.\]
ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]\]
Given matrix and Vector:
\[A = \left[ {\begin{array}{*{20}{r}}1&0&{ - 2}\\{ - 2}&1&6\\3&{ - 2}&{ - 5}\end{array}} \right],{\bf{b}} = \left[ {\begin{array}{*{20}{r}}{ - 1}\\7\\{ - 3}\end{array}} \right]\]Given Transformation:\[T:{R^3} \to {R^2}::\,T\left( {\bf{x}} \right) = A{\bf{x}}\]We have to solve the system of equations given by: \[T\left( {\bf{x}} \right) = {\bf{b}}\]
Step 2: The system of Equations
\[\left[ {\begin{array}{*{20}{r}}1&0&{ - 2}\\{ - 2}&1&6\\3&{ - 2}&{ - 5}\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{r}}{ - 1}\\7\\{ - 3}\end{array}} \right]\]
Step 3: The corresponding Augmented matrix form
\[M = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\{ - 2}&1&6&7\\3&{ - 2}&{ - 5}&{ - 3}\end{array}} \right]\]
Step 4: Convert to row reduced echelon form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\{ - 2}&1&6&7\\3&{ - 2}&{ - 5}&{ - 3}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&{ - 2}&1&0\end{array}} \right]::\,{R_3} = {R_3} - 3{R_1};\,\,{R_2} = {R_2} + 2{R_1}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&0&5&{10}\end{array}} \right]::\,{R_3} = {R_3} + 2{R_2}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&2&5\\0&0&1&2\end{array}} \right]::\,{R_3} = {R_3}/5\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 2}&{ - 1}\\0&1&0&1\\0&0&1&2\end{array}} \right]::\,{R_3} = {R_2} - 2{R_3}\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&3\\0&1&0&1\\0&0&1&2\end{array}} \right]::\,{R_1} = {R_1} + 2{R_3}\end{array}\]
Step 5: The System of Equations
Since there are three variables and three equations:\[\left\{ \begin{array}{l}{x_1} = 3\\{x_2} = 1\\{x_3} = 2\end{array} \right.\]
ANSWER
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}}3\\1\\2\end{array}} \right]\]