Linear Algebra and Its Applications, 5th Edition

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 26E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.

Problem 26E

Chapter:
Problem:
Construct three different augmented matrices for linear systems whose solution set is x1 = −2, x2 = 1, x3 = 0.

Step-by-Step Solution

Given Information
We are given with a solution set: \[{x_1} = - 2,\,\,{x_2} = 1,\,\,{x_3} = 0\] We have to find the find some augmented matrices that have the solution matrix given by above solution set \\

Step-1: The Augmented matrix
The most basic augmented matrix is the row-reduced echelon form as shown below: \[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 0&1&0&1\\ 0&0&1&0 \end{array}} \right]\,\, \Rightarrow \,\,\,\left\{ {\begin{array}{*{20}{l}} {{x_1} = - 2}\\ {{x_2} = 1}\\ {{x_3} = 0} \end{array}} \right\}\]

\[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 0&1&0&1\\ 0&0&1&0 \end{array}} \right]\]


Step-2: The Augmented matrix-2
Apply elementary row operation (Add 3 times row-1 to row-2) to above matrix to get another form of augmented matrix. \[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 3&1&0&{ - 5}\\ 0&0&1&0 \end{array}} \right] \Rightarrow \,\,\,\left\{ {\begin{array}{*{20}{l}} {{x_1} = - 2}\\ {3{x_1} + {x_2} = - 5}\\ {{x_3} = 0} \end{array}} \right\}\]

\[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 3&1&0&{ - 5}\\ 0&0&1&0 \end{array}} \right]\]


Step-3: The Augmented matrix-3
Apply elementary row operation (Add 3 times row-1 to row-3) to above matrix to get another form of augmented matrix. \[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 3&1&0&{ - 5}\\ 3&0&1&{ - 6} \end{array}} \right] \Rightarrow \,\,\,\left\{ {\begin{array}{*{20}{l}} {{x_1} = - 2}\\ {3{x_1} + {x_2} = - 5}\\ {3{x_1} + {x_3} = - 6} \end{array}} \right\}\]

\[\left[ {\begin{array}{*{20}{c}} 1&0&0&{ - 2}\\ 3&1&0&{ - 5}\\ 3&0&1&{ - 6} \end{array}} \right]\]