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 7E from Chapter 1.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
In Exercises 5-8, determine if the columns of the matrix form a linearly independent set. Justify each answer...
Step-by-Step Solution
Step 1
Given Matrix\[A = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\]There are four column vectors in above matrix. We have to check whether the vectors are independent or not. As there are four columns, we will have 4 variables and 3 equations (number of rows).
Step 2: The matrix form of equations
\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\]
Step 3: The Augmented Matrix
\[M = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\{ - 2}&{ - 7}&5&1&0\\{ - 4}&{ - 5}&7&5&0\end{array}} \right]\]
Step 4: Row Reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\{ - 2}&{ - 7}&5&1&0\\{ - 4}&{ - 5}&7&5&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&{11}&{ - 5}&5&0\end{array}} \right]::\,{R_3} = {R_3} + 4{R_1};\,\,\,\,{R_2} = {R_2} + 2{R_1}\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&0&6&{ - 6}&0\end{array}} \right]::\,{R_3} = {R_3} - 11{R_1};\,\,\,\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_3} = {R_3}/6;\,\,\,\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_2} = {R_2} + {R_3}\,;\,\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_1} = {R_1} - 4{R_2}\,;\,\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 3}&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_1} = {R_1} + 3{R_3}\,;\,\end{array}\]As there are 4 variables and three equations, consider one variable ($x_4$) as free variable. The general solution is:\[\left\{ \begin{array}{l}{x_1} - 3{x_4} = 0\\{x_2} = 0\\{x_3} - {x_4} = 0\end{array} \right.\]
ANSWER
As we have one free variable, the columns are linearly dependent and system has non trivial solution
Given Matrix\[A = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\]There are four column vectors in above matrix. We have to check whether the vectors are independent or not. As there are four columns, we will have 4 variables and 3 equations (number of rows).
Step 2: The matrix form of equations
\[\begin{array}{l}A{\bf{x}} = 0\\\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\]
Step 3: The Augmented Matrix
\[M = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\{ - 2}&{ - 7}&5&1&0\\{ - 4}&{ - 5}&7&5&0\end{array}} \right]\]
Step 4: Row Reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\{ - 2}&{ - 7}&5&1&0\\{ - 4}&{ - 5}&7&5&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&{11}&{ - 5}&5&0\end{array}} \right]::\,{R_3} = {R_3} + 4{R_1};\,\,\,\,{R_2} = {R_2} + 2{R_1}\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&0&6&{ - 6}&0\end{array}} \right]::\,{R_3} = {R_3} - 11{R_1};\,\,\,\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_3} = {R_3}/6;\,\,\,\\ = \left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_2} = {R_2} + {R_3}\,;\,\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 3}&0&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_1} = {R_1} - 4{R_2}\,;\,\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&{ - 3}&0\\0&1&0&0&0\\0&0&1&{ - 1}&0\end{array}} \right]::\,{R_1} = {R_1} + 3{R_3}\,;\,\end{array}\]As there are 4 variables and three equations, consider one variable ($x_4$) as free variable. The general solution is:\[\left\{ \begin{array}{l}{x_1} - 3{x_4} = 0\\{x_2} = 0\\{x_3} - {x_4} = 0\end{array} \right.\]
ANSWER
As we have one free variable, the columns are linearly dependent and system has non trivial solution