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 5E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
Consider each matrix in Exercises 5 and 6 as the augmented matrix of a linear system. State in words the next two elementary row operations that should be performed in the process of solving the
Step-by-Step Solution
Given Information
We are given with an augmented matrix. \[M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\]We have to find next two operations to solve the system of equations.
Step-1: First Operation
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 3{R_3}} \right\}\end{array}\]The first operation is
Replace Row-2 with sum of 3 times Row-3 and Row-2
Step-2: Second Operation
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&0&0&{ - 3}\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\,\,::\,\,\left\{ {{R_1} = {R_1} - 5{R_3}} \right\}\end{array}\]The Second operation is
Replace Row-1 with sum of -5 times Row-3 and Row-1
We are given with an augmented matrix. \[M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\]We have to find next two operations to solve the system of equations.
Step-1: First Operation
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 3{R_3}} \right\}\end{array}\]The first operation is
Replace Row-2 with sum of 3 times Row-3 and Row-2
Step-2: Second Operation
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&{ - 3}&0&6\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&5&0&7\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} + 3{R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&0&0&{ - 3}\\0&1&0&0&{12}\\0&0&1&0&2\\0&0&0&1&{ - 5}\end{array}} \right]\,\,::\,\,\left\{ {{R_1} = {R_1} - 5{R_3}} \right\}\end{array}\]The Second operation is
Replace Row-1 with sum of -5 times Row-3 and Row-1