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 1.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
In Exercises 3-6, with T defined by T(x) = Ax, find a vector x whose image under T is b, and determine whether x is unique.
Step-by-Step Solution
Step 1
Given matrix and Vector:
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}\\{ - 3}&7&5\end{array}} \right],{\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 2}\end{array}} \right]\]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
\[\begin{array}{l}A{\bf{x}} = b\\\\\left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}\\{ - 3}&7&5\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 2}\end{array}} \right]\end{array}\]
Step 3: The corresponding Augmented matrix form
\[M = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\{ - 3}&7&5&{ - 2}\end{array}} \right]\]
Step 4: Convert to reduced echelon form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\{ - 3}&7&5&{ - 2}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&{ - 8}&{ - 16}&{ - 8}\end{array}} \right]:\,\,\left\{ {{R_2} = {R_2} + 3{R_1}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&1&2&1\end{array}} \right]:\,\,\left\{ {{R_2} = - \frac{{{R_2}}}{8}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}{\left\langle 1 \right\rangle }&0&3&3\\0&{\left\langle 1 \right\rangle }&2&1\end{array}} \right]:\,\,\left\{ {{R_1} = {R_1} + 5{R_2}} \right\}\end{array}\]
Step 5: The System of Equations
Since there are three variables and two equations, one variable is considered as free. Let $x_3$ be free:\[\begin{array}{l}{x_1} + 3{x_3} = 3 \Rightarrow {x_1} = 3 - 3{x_3}\\\\{x_2} + 2{x_3} = 1 \Rightarrow {x_2} = 1 - 2{x_3}\end{array}\]
Step 5: The Solution
The solution in Matrix form is given by: \[{\bf{x}} = \left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 3{x_3}}\\{1 - 2{x_3}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}3\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\end{array}} \right]\]
ANSWER
\[{\rm{Vector}}\,\,{\bf{x}}\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{Unique}}\]
Given matrix and Vector:
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}\\{ - 3}&7&5\end{array}} \right],{\bf{b}} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 2}\end{array}} \right]\]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
\[\begin{array}{l}A{\bf{x}} = b\\\\\left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}\\{ - 3}&7&5\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 2}\end{array}} \right]\end{array}\]
Step 3: The corresponding Augmented matrix form
\[M = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\{ - 3}&7&5&{ - 2}\end{array}} \right]\]
Step 4: Convert to reduced echelon form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\{ - 3}&7&5&{ - 2}\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&{ - 8}&{ - 16}&{ - 8}\end{array}} \right]:\,\,\left\{ {{R_2} = {R_2} + 3{R_1}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 5}&{ - 7}&{ - 2}\\0&1&2&1\end{array}} \right]:\,\,\left\{ {{R_2} = - \frac{{{R_2}}}{8}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}{\left\langle 1 \right\rangle }&0&3&3\\0&{\left\langle 1 \right\rangle }&2&1\end{array}} \right]:\,\,\left\{ {{R_1} = {R_1} + 5{R_2}} \right\}\end{array}\]
Step 5: The System of Equations
Since there are three variables and two equations, one variable is considered as free. Let $x_3$ be free:\[\begin{array}{l}{x_1} + 3{x_3} = 3 \Rightarrow {x_1} = 3 - 3{x_3}\\\\{x_2} + 2{x_3} = 1 \Rightarrow {x_2} = 1 - 2{x_3}\end{array}\]
Step 5: The Solution
The solution in Matrix form is given by: \[{\bf{x}} = \left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{3 - 3{x_3}}\\{1 - 2{x_3}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}3\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 2}\\1\end{array}} \right]\]
ANSWER
\[{\rm{Vector}}\,\,{\bf{x}}\,\,{\rm{is}}\,\,{\rm{not}}\,\,{\rm{Unique}}\]