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 4E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 4E
Chapter:
Problem:
Find an explicit description of Nul A by listing vectors that span the null space.
Step-by-Step Solution
Given Information
We are given with following matrix: \[ A = \left[ \begin{array} { c c c c } { 1 } & { - 6 } & { 4 } & { 0 } \\ { 0 } & { 0 } & { 2 } & { 0 } \end{array} \right] \] We have to find an explicit description of Nul A by listing vectors that span the null space. The null space is the set of all homogeneous solutions of $Ax=0$.
Step-1: General solution of $Ax=0$
Matrix form of equations: \[\begin{array}{l} A{\bf{x}} = 0\\ \left[ {\begin{array}{*{20}{c}} 1&{ - 6}&4&0\\ 0&0&2&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right] \end{array}\] System of equations: \[\begin{array}{l} {x_1} - 6{x_2} + 4{x_3} = 0\\ 2{x_3} = 0 \end{array}\] Consider $x_2$ and $x_4$ as free, \[\begin{array}{l} {x_1} = 6{x_2} - 4{x_3}\\ {x_2} = s\\ {x_3} = 9\\ {x_4} = t \end{array}\] So, solution in parametric form is: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {6{x_2}}\\ {{x_2}}\\ 0\\ t \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0\\ 0 \end{array}} \right]{s} + \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 1 \end{array}} \right]t \end{array}\] Therefore, the Nul space is:
\[Nul\,A = \left\{ {\left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 1 \end{array}} \right]} \right\}\]
We are given with following matrix: \[ A = \left[ \begin{array} { c c c c } { 1 } & { - 6 } & { 4 } & { 0 } \\ { 0 } & { 0 } & { 2 } & { 0 } \end{array} \right] \] We have to find an explicit description of Nul A by listing vectors that span the null space. The null space is the set of all homogeneous solutions of $Ax=0$.
Step-1: General solution of $Ax=0$
Matrix form of equations: \[\begin{array}{l} A{\bf{x}} = 0\\ \left[ {\begin{array}{*{20}{c}} 1&{ - 6}&4&0\\ 0&0&2&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} 0\\ 0 \end{array}} \right] \end{array}\] System of equations: \[\begin{array}{l} {x_1} - 6{x_2} + 4{x_3} = 0\\ 2{x_3} = 0 \end{array}\] Consider $x_2$ and $x_4$ as free, \[\begin{array}{l} {x_1} = 6{x_2} - 4{x_3}\\ {x_2} = s\\ {x_3} = 9\\ {x_4} = t \end{array}\] So, solution in parametric form is: \[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {6{x_2}}\\ {{x_2}}\\ 0\\ t \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0\\ 0 \end{array}} \right]{s} + \left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 1 \end{array}} \right]t \end{array}\] Therefore, the Nul space is:
\[Nul\,A = \left\{ {\left[ {\begin{array}{*{20}{l}} 6\\ 1\\ 0\\ 0 \end{array}} \right],\left[ {\begin{array}{*{20}{l}} 0\\ 0\\ 0\\ 1 \end{array}} \right]} \right\}\]