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 6E from Chapter 1.SE from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 6E
Chapter:
Problem:
Consider the problem of determining whether the following system of equations is consistent: a. Define appropriate vectors, and restate the problem in terms of linear combinations. Then solve that problem. b. Define an appropriate matrix, and restate the problem using the phrase “columns of A.” c. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.
Step-by-Step Solution
Given Information
We are given with following system of equations: \[ \begin{array} { l } { 4 x _ { 1 } - 2 x _ { 2 } + 7 x _ { 3 } = - 5 } \\ { 8 x _ { 1 } - 3 x _ { 2 } + 10 x _ { 3 } = - 3 } \end{array} \] We have to restate the problem in various forms and check if the system is consistent.
Step-1:
The Matrix form of system of equations: \[ \left[ \begin{array} { c c c } { 4 } & { - 2 } & { 7 } \\ { 8 } & { - 3 } & { 10 } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] \] Let us consider the vectors \[{v_1} = \left[ {\begin{array}{*{20}{l}} 4\\ 8 \end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{l}} { - 2}\\ { - 3} \end{array}} \right],{v_3} = \left[ {\begin{array}{*{20}{c}} 7\\ {10} \end{array}} \right]b = \left[ {\begin{array}{*{20}{l}} { - 5}\\ { - 3} \end{array}} \right]\] Row-Reduced augmented Matrix form: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7&{ - 5}\\ 8&{ - 3}&{10}&{ - 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7&{ - 5}\\ 0&1&{ - 4}&7 \end{array}} \right]::\left\{ {{R_2} \to {R_2} - 2{R_1}} \right\} \end{array}\] Since rank of matrix A is 2 (less than number of variables), hence the system is consistent. Hence, b is span ${v_1, v_2, v_3}$.
Step-2: (b)
Write the system in vector form: \[ \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] = x _ { 1 } \left[ \begin{array} { c } { 4 } \\ { 8 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { c } { - 2 } \\ { - 3 } \end{array} \right] + x _ { 3 } \left[ \begin{array} { c } { 7 } \\ { 10 } \end{array} \right] \] The Row-Reduced form of Augmented matrix is same as above: \[ \left[ \begin{array} { c c c c } { 4 } & { - 2 } & { 7 } & { - 5 } \\ { 0 } & { 1 } & { - 4 } & { 7 } \end{array} \right] \] Since rank of matrix A is 2 (less than number of variables), hence the system is consistent. Therefore, b is linear combination of columns of A.
Step-3: (c)
Write in Transformation form: \[\begin{array}{l} T({\bf{x}}) = A{\bf{x}}\\ T\left( {\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7\\ 8&{ - 3}&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] \end{array}\] Check if $b$ is in range of T \[ \begin{aligned} T \left[ \dfrac { 9 } { 4 } \right] & \\ T & = \left[ \begin{array} { c c c } { 4 } & { - 2 } & { 7 } \\ { 8 } & { - 3 } & { 10 } \end{array} \right] \left[ \begin{array} { c } { \dfrac { 9 } { 4 } } \\ { 7 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 9 - 14 + 0 } \\ { 18 - 21 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] \end{aligned} \] Therefore,
The vector b is in the range of T.
We are given with following system of equations: \[ \begin{array} { l } { 4 x _ { 1 } - 2 x _ { 2 } + 7 x _ { 3 } = - 5 } \\ { 8 x _ { 1 } - 3 x _ { 2 } + 10 x _ { 3 } = - 3 } \end{array} \] We have to restate the problem in various forms and check if the system is consistent.
Step-1:
The Matrix form of system of equations: \[ \left[ \begin{array} { c c c } { 4 } & { - 2 } & { 7 } \\ { 8 } & { - 3 } & { 10 } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] \] Let us consider the vectors \[{v_1} = \left[ {\begin{array}{*{20}{l}} 4\\ 8 \end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{l}} { - 2}\\ { - 3} \end{array}} \right],{v_3} = \left[ {\begin{array}{*{20}{c}} 7\\ {10} \end{array}} \right]b = \left[ {\begin{array}{*{20}{l}} { - 5}\\ { - 3} \end{array}} \right]\] Row-Reduced augmented Matrix form: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7&{ - 5}\\ 8&{ - 3}&{10}&{ - 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7&{ - 5}\\ 0&1&{ - 4}&7 \end{array}} \right]::\left\{ {{R_2} \to {R_2} - 2{R_1}} \right\} \end{array}\] Since rank of matrix A is 2 (less than number of variables), hence the system is consistent. Hence, b is span ${v_1, v_2, v_3}$.
Step-2: (b)
Write the system in vector form: \[ \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] = x _ { 1 } \left[ \begin{array} { c } { 4 } \\ { 8 } \end{array} \right] + x _ { 2 } \left[ \begin{array} { c } { - 2 } \\ { - 3 } \end{array} \right] + x _ { 3 } \left[ \begin{array} { c } { 7 } \\ { 10 } \end{array} \right] \] The Row-Reduced form of Augmented matrix is same as above: \[ \left[ \begin{array} { c c c c } { 4 } & { - 2 } & { 7 } & { - 5 } \\ { 0 } & { 1 } & { - 4 } & { 7 } \end{array} \right] \] Since rank of matrix A is 2 (less than number of variables), hence the system is consistent. Therefore, b is linear combination of columns of A.
Step-3: (c)
Write in Transformation form: \[\begin{array}{l} T({\bf{x}}) = A{\bf{x}}\\ T\left( {\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}} 4&{ - 2}&7\\ 8&{ - 3}&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{l}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] \end{array}\] Check if $b$ is in range of T \[ \begin{aligned} T \left[ \dfrac { 9 } { 4 } \right] & \\ T & = \left[ \begin{array} { c c c } { 4 } & { - 2 } & { 7 } \\ { 8 } & { - 3 } & { 10 } \end{array} \right] \left[ \begin{array} { c } { \dfrac { 9 } { 4 } } \\ { 7 } \\ { 0 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 9 - 14 + 0 } \\ { 18 - 21 } \end{array} \right] \\ & = \left[ \begin{array} { c } { - 5 } \\ { - 3 } \end{array} \right] \end{aligned} \] Therefore,
The vector b is in the range of T.