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 20E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 20E
Chapter:
Problem:
Exercises 17–20 refer to the matrices A and B below. Make appropriate calculations that justify your answers and mention an appropriate theorem. Can every vector in ℝ4 be written as a linear combination of the columns of the matrix B above? Do the columns of B span ℝ3?
Step-by-Step Solution
Given Information
We are given with following matrices \[ A = \left[ \begin{array} { r r r r } { 1 } & { 3 } & { 0 } & { 3 } \\ { - 1 } & { - 1 } & { - 1 } & { 1 } \\ { 0 } & { - 4 } & { 2 } & { - 8 } \\ { 2 } & { 0 } & { 3 } & { - 1 } \end{array} \right] \quad B = \left[ \begin{array} { c c c c } { 1 } & { 4 } & { 1 } & { 2 } \\ { 0 } & { 1 } & { 3 } & { - 4 } \\ { 0 } & { 2 } & { 6 } & { 7 } \\ { 2 } & { 9 } & { 5 } & { - 7 } \end{array} \right] \] We have to tell if the every vector in $R^4$ be written as a linear combination of the matrix B? Also, we have to tell if the columns of A span $R^4$?
Step-1: Row-Reduce the matrix B
\[\begin{array}{l} B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&{ - 1}&{ - 1}&5\\ 0&{ - 2}&{ - 2}&3 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&0&0&0\\ 0&0&0&{ - 7} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] Since not all the rows have a pivot element, hence columns of B do not span $R^4$.
Step-2:
We have to tell if the equation $Bx=y$ have a solution for each y in $R^4$. Let, \[ y = \left[ \begin{array} { l } { b _ { 1 } } \\ { b _ { 2 } } \\ { b _ { 3 } } \\ { b _ { 4 } } \end{array} \right] \]
Step-3: The augmented matrix
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 1&2&{ - 3}&7&{{b_3}}\\ { - 2}&{ - 8}&2&{ - 1}&{{b_4}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&1\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + {b_4}} \end{array}} \right]::\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&1\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_2} + {b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + 2{b_2} + {b_4}} \end{array}} \right]::\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] From the last, row, if $b _ { 2 } + b _ { 3 } - b _ { 1 }$ is nonzero, then the system does not have a solution. Therefore,
The system does not have a solution for each y in $R^4$
We are given with following matrices \[ A = \left[ \begin{array} { r r r r } { 1 } & { 3 } & { 0 } & { 3 } \\ { - 1 } & { - 1 } & { - 1 } & { 1 } \\ { 0 } & { - 4 } & { 2 } & { - 8 } \\ { 2 } & { 0 } & { 3 } & { - 1 } \end{array} \right] \quad B = \left[ \begin{array} { c c c c } { 1 } & { 4 } & { 1 } & { 2 } \\ { 0 } & { 1 } & { 3 } & { - 4 } \\ { 0 } & { 2 } & { 6 } & { 7 } \\ { 2 } & { 9 } & { 5 } & { - 7 } \end{array} \right] \] We have to tell if the every vector in $R^4$ be written as a linear combination of the matrix B? Also, we have to tell if the columns of A span $R^4$?
Step-1: Row-Reduce the matrix B
\[\begin{array}{l} B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&{ - 1}&{ - 1}&5\\ 0&{ - 2}&{ - 2}&3 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&0&0&0\\ 0&0&0&{ - 7} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] Since not all the rows have a pivot element, hence columns of B do not span $R^4$.
Step-2:
We have to tell if the equation $Bx=y$ have a solution for each y in $R^4$. Let, \[ y = \left[ \begin{array} { l } { b _ { 1 } } \\ { b _ { 2 } } \\ { b _ { 3 } } \\ { b _ { 4 } } \end{array} \right] \]
Step-3: The augmented matrix
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 1&2&{ - 3}&7&{{b_3}}\\ { - 2}&{ - 8}&2&{ - 1}&{{b_4}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&1\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + {b_4}} \end{array}} \right]::\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&1\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_2} + {b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + 2{b_2} + {b_4}} \end{array}} \right]::\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] From the last, row, if $b _ { 2 } + b _ { 3 } - b _ { 1 }$ is nonzero, then the system does not have a solution. Therefore,
The system does not have a solution for each y in $R^4$