Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 32E from Chapter 1.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 32E
Chapter:
Problem:
Given A = observe that the first column plus twice the second column equals the third column. Find a nontrivial solution of Ax = 0.
Step-by-Step Solution
Given Information
We are given with a matrix A\[A = \left[ {\begin{array}{*{20}{c}}4&1&6\\{ - 7}&5&3\\9&{ - 3}&3\end{array}} \right]\]We have to find the nontrivial solution of $Ax=0$
Step-1:
Let,\[A = \left[ {\begin{array}{*{20}{c}}4&1&6\\{ - 7}&5&3\\9&{ - 3}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right]\]We can notice that third column can be obtained by adding two times the second column with first column. \[\begin{array}{l}6 = 4 + 2 \times 1\\3 = - 7 + 2 \times 5\\3 = 9 + 2 \times \left( { - 3} \right)\end{array}\]So, \[{a_3} = {a_1} + 2{a_2}\]Therefore, one of the nontrivial solution is
\[{\bf{x}} = \left( {1,2, - 1} \right)\]
We are given with a matrix A\[A = \left[ {\begin{array}{*{20}{c}}4&1&6\\{ - 7}&5&3\\9&{ - 3}&3\end{array}} \right]\]We have to find the nontrivial solution of $Ax=0$
Step-1:
Let,\[A = \left[ {\begin{array}{*{20}{c}}4&1&6\\{ - 7}&5&3\\9&{ - 3}&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{a_1}}&{{a_2}}&{{a_3}}\end{array}} \right]\]We can notice that third column can be obtained by adding two times the second column with first column. \[\begin{array}{l}6 = 4 + 2 \times 1\\3 = - 7 + 2 \times 5\\3 = 9 + 2 \times \left( { - 3} \right)\end{array}\]So, \[{a_3} = {a_1} + 2{a_2}\]Therefore, one of the nontrivial solution is
\[{\bf{x}} = \left( {1,2, - 1} \right)\]