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 31E from Chapter 1.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 31E
Chapter:
Problem:
Exercises 31 and 32 should be solved without performing row operations. [Hint: Write Ax = 0 as a vector equation.] Given observe that the third column is the sum of the first two columns. Find a nontrivial solution of Ax = 0.
Step-by-Step Solution
Given Information
We are given with a matrix \[A = \left[ {\begin{array}{*{20}{r}} 2&3&5\\ { - 5}&1&{ - 4}\\ { - 3}&{ - 1}&{ - 4}\\ 1&0&1 \end{array}} \right]\] It is given that the third column is sum of first two columns. We have to find a nontrivial solution of $Ax=0$.
Step-1:
If the columns are represented by $a_1$, $a_2$ and $a_3$, then using the given relationship, we can write \[\begin{array}{l} {{\bf{a}}_{\bf{3}}} = {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}}\\ {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}} = {{\bf{a}}_{\bf{3}}}\\ {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}} - {{\bf{a}}_{\bf{3}}} = 0\\ \left( 1 \right){{\bf{a}}_{\bf{1}}} + \left( 1 \right){{\bf{a}}_{\bf{2}}} + \left( { - 1} \right){{\bf{a}}_{\bf{3}}} = 0 \end{array}\] The above form can be converted into matrix form as shown below. \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{a}}_1}\,\,\,{{\bf{a}}_2}\,\,\,{{\bf{a}}_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right] = {\bf{0}}\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{a}}_1}\,\,\,{{\bf{a}}_2}\,\,\,{{\bf{a}}_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = {\bf{0}}\\ A{\bf{x}} = {\bf{0}} \end{array}\] Therefore, the solution of the system is:
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right]\]
We are given with a matrix \[A = \left[ {\begin{array}{*{20}{r}} 2&3&5\\ { - 5}&1&{ - 4}\\ { - 3}&{ - 1}&{ - 4}\\ 1&0&1 \end{array}} \right]\] It is given that the third column is sum of first two columns. We have to find a nontrivial solution of $Ax=0$.
Step-1:
If the columns are represented by $a_1$, $a_2$ and $a_3$, then using the given relationship, we can write \[\begin{array}{l} {{\bf{a}}_{\bf{3}}} = {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}}\\ {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}} = {{\bf{a}}_{\bf{3}}}\\ {{\bf{a}}_{\bf{1}}} + {{\bf{a}}_{\bf{2}}} - {{\bf{a}}_{\bf{3}}} = 0\\ \left( 1 \right){{\bf{a}}_{\bf{1}}} + \left( 1 \right){{\bf{a}}_{\bf{2}}} + \left( { - 1} \right){{\bf{a}}_{\bf{3}}} = 0 \end{array}\] The above form can be converted into matrix form as shown below. \[\begin{array}{l} \left[ {\begin{array}{*{20}{l}} {{{\bf{a}}_1}\,\,\,{{\bf{a}}_2}\,\,\,{{\bf{a}}_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right] = {\bf{0}}\\ \left[ {\begin{array}{*{20}{l}} {{{\bf{a}}_1}\,\,\,{{\bf{a}}_2}\,\,\,{{\bf{a}}_3}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}} \end{array}} \right] = {\bf{0}}\\ A{\bf{x}} = {\bf{0}} \end{array}\] Therefore, the solution of the system is:
\[{\bf{x}} = \left[ {\begin{array}{*{20}{c}} 1\\ 1\\ { - 1} \end{array}} \right]\]