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 9E from Chapter 2.9 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 9E
Chapter:
Problem:
Exercises display a matrix A and an echelon form of A. Find bases for Col A and Nul A, and then state the dimensions of these subspaces.
Step-by-Step Solution
Given Information
We are given with a matrix A \[ A = \left[ \begin{array} { r r r r } { 1 } & { - 3 } & { 2 } & { - 4 } \\ { - 3 } & { 9 } & { - 1 } & { 5 } \\ { 2 } & { - 6 } & { 4 } & { - 3 } \\ { - 4 } & { 12 } & { 2 } & { 7 } \end{array} \right] \] We are also given with its echelon form. \[A = \left[ {\begin{array}{*{20}{r}} 1&{ - 3}&2&{ - 4}\\ 0&0&5&{ - 7}\\ 0&0&0&5\\ 0&0&0&0 \end{array}} \right]\] We have to find the bases for Col A and Nul A, and then state the dimensions of these subspaces.
Step-1: Basis for Col A
Consider the echelon form of the matrix, we can see that three columns (1,3,4) have pivot elements. Therefore, the column space is given by the corresponding columns of the matrix A. \[{\rm{Col}}\,\,A\,\, = \,\left\{ {\left[ {\begin{array}{*{20}{c}} 1\\ { - 3}\\ 2\\ { - 4} \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 4\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} { - 4}\\ 5\\ { - 3}\\ 7 \end{array}} \right]} \right\}\]
Step-2: Basis for Nul A
Write the augmented matrix and row-reduce it \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&{ - 4}&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&5&0\\ 0&0&0&0&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&{ - 4}&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_3} = \dfrac{{{R_3}}}{5}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&0&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 4{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&0&0\\ 0&0&5&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 7{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - 2{R_2}} \right\} \end{array}\]
Step-3: General solution
System of equations are \[ \begin{aligned} x _ { 1 } - 3 x _ { 2 } & = 0 \\ x _ { 3 } & = 0 \\ x _ { 4 } & = 0 \end{aligned} \] By considering $x_2$ as free, the general solution is \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3{x_2}}\\ {{x_2}}\\ 0\\ 0 \end{array}} \right] = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right]\] Therefore the Nul A is given by:
\[{\rm{Nul}}\,\,A\, = \,\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right]\]
We are given with a matrix A \[ A = \left[ \begin{array} { r r r r } { 1 } & { - 3 } & { 2 } & { - 4 } \\ { - 3 } & { 9 } & { - 1 } & { 5 } \\ { 2 } & { - 6 } & { 4 } & { - 3 } \\ { - 4 } & { 12 } & { 2 } & { 7 } \end{array} \right] \] We are also given with its echelon form. \[A = \left[ {\begin{array}{*{20}{r}} 1&{ - 3}&2&{ - 4}\\ 0&0&5&{ - 7}\\ 0&0&0&5\\ 0&0&0&0 \end{array}} \right]\] We have to find the bases for Col A and Nul A, and then state the dimensions of these subspaces.
Step-1: Basis for Col A
Consider the echelon form of the matrix, we can see that three columns (1,3,4) have pivot elements. Therefore, the column space is given by the corresponding columns of the matrix A. \[{\rm{Col}}\,\,A\,\, = \,\left\{ {\left[ {\begin{array}{*{20}{c}} 1\\ { - 3}\\ 2\\ { - 4} \end{array}} \right],\left[ {\begin{array}{*{20}{c}} 2\\ { - 1}\\ 4\\ 2 \end{array}} \right],\left[ {\begin{array}{*{20}{c}} { - 4}\\ 5\\ { - 3}\\ 7 \end{array}} \right]} \right\}\]
Step-2: Basis for Nul A
Write the augmented matrix and row-reduce it \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&{ - 4}&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&5&0\\ 0&0&0&0&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&{ - 4}&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_3} = \dfrac{{{R_3}}}{5}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&0&0\\ 0&0&5&{ - 7}&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 4{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&2&0&0\\ 0&0&5&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 7{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 3}&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0 \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - 2{R_2}} \right\} \end{array}\]
Step-3: General solution
System of equations are \[ \begin{aligned} x _ { 1 } - 3 x _ { 2 } & = 0 \\ x _ { 3 } & = 0 \\ x _ { 4 } & = 0 \end{aligned} \] By considering $x_2$ as free, the general solution is \[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {3{x_2}}\\ {{x_2}}\\ 0\\ 0 \end{array}} \right] = {x_2}\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right]\] Therefore the Nul A is given by:
\[{\rm{Nul}}\,\,A\, = \,\left[ {\begin{array}{*{20}{l}} 3\\ 1\\ 0\\ 0 \end{array}} \right]\]