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 18E from Chapter 1.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 18E
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. Do the columns of B span ℝ4? Does the equation Bx = y have a solution for each y in ℝ4?
Step-by-Step Solution
Given Information
We are given with a matrix: \[B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\] We have to find whether the columns of matrix B span $R^4$. Also we have to find whether, the equation $Ax=b$ has a solution for each b in $R^4$.
Step-1: Row-Reduced echelon form of the matrix
\[\begin{array}{l} B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&{ - 1}&{ - 1}&5\\ 0&{ - 2}&{ - 2}&3 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&0&0&0\\ 0&0&0&{ - 7} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] Not all the rows have a pivot element, hence the the columns of B do not span $R^4$
Step-2: The augmented matrix
Let us consider an arbitrary vector $y$ in $R^4$. \[y = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}}\\ {{b_3}}\\ {{b_4}} \end{array}} \right]\] The augmented matrix for $Bx=y$ \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 1&2&{ - 3}&7&{{b_3}}\\ { - 2}&{ - 8}&2&{ - 1}&{{b_4}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + {b_4}} \end{array}} \right]::\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&0&0&0&{{b_2} + {b_3} - {b_1}}\\ 0&0&0&{ - 7}&{2{b_1} + 2{b_2} + {b_4}} \end{array}} \right]::\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] If the term $b_2 + b_3 - b_1$ is nonzero, the third row is never true, Therefore,
The matrix equation does not have a solution for each $y$ in $R^4$.
We are given with a matrix: \[B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\] We have to find whether the columns of matrix B span $R^4$. Also we have to find whether, the equation $Ax=b$ has a solution for each b in $R^4$.
Step-1: Row-Reduced echelon form of the matrix
\[\begin{array}{l} B = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 1&2&{ - 3}&7\\ { - 2}&{ - 8}&2&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&{ - 1}&{ - 1}&5\\ 0&{ - 2}&{ - 2}&3 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&3&{ - 2}&2\\ 0&1&1&{ - 5}\\ 0&0&0&0\\ 0&0&0&{ - 7} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] Not all the rows have a pivot element, hence the the columns of B do not span $R^4$
Step-2: The augmented matrix
Let us consider an arbitrary vector $y$ in $R^4$. \[y = \left[ {\begin{array}{*{20}{c}} {{b_1}}\\ {{b_2}}\\ {{b_3}}\\ {{b_4}} \end{array}} \right]\] The augmented matrix for $Bx=y$ \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 1&2&{ - 3}&7&{{b_3}}\\ { - 2}&{ - 8}&2&{ - 1}&{{b_4}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&{ - 1}&{ - 1}&5&{{b_3} - {b_1}}\\ 0&{ - 2}&{ - 2}&3&{2{b_1} + {b_4}} \end{array}} \right]::\left\{ \begin{array}{l} {R_3} = {R_3} - {R_1}\\ {R_4} = {R_4} + 2{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{r}} 1&3&{ - 2}&2&{{b_1}}\\ 0&1&1&{ - 5}&{{b_2}}\\ 0&0&0&0&{{b_2} + {b_3} - {b_1}}\\ 0&0&0&{ - 7}&{2{b_1} + 2{b_2} + {b_4}} \end{array}} \right]::\left\{ \begin{array}{l} {R_3} = {R_3} + {R_2}\\ {R_4} = {R_4} + 2{R_2} \end{array} \right\} \end{array}\] If the term $b_2 + b_3 - b_1$ is nonzero, the third row is never true, Therefore,
The matrix equation does not have a solution for each $y$ in $R^4$.