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 4E from Chapter 1.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 4E
Chapter:
Problem:
Row reduce the matrices to reduced echelon form. Circle the pivot positions in the final matrix and in the original matrix, and list the pivot columns.
Step-by-Step Solution
Given Information
We are given with a matrix A\[A = \left[ {\begin{array}{*{20}{l}}1&3&5&7\\3&5&7&9\\5&7&9&1\end{array}} \right]\]We have to Row reduce the matrix to reduced echelon form and find the pivot element.
Step-1: Row-Reduced Echelon form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}1&3&5&7\\3&5&7&9\\5&7&9&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&{ - 8}&{ - 16}&{ - 34}\end{array}} \right]::\left\{ \begin{array}{l}{R_2} = {R_2} - 3{R_1}\\{R_3} = {R_3} - 5{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&{ - 4}&{ - 8}&{ - 17}\end{array}} \right]::\left\{ {{R_3} = \dfrac{{{R_3}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&0&0&{ - 5}\end{array}} \right]::\left\{ {{R_3} = {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&1&2&3\\0&0&0&1\end{array}} \right]::\left\{ \begin{array}{l}{R_3} = - \dfrac{{{R_3}}}{5}\\{R_2} = - \dfrac{{{R_2}}}{4}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&0\\0&1&2&0\\0&0&0&1\end{array}} \right]::\left\{ \begin{array}{l}{R_2} = {R_2} - 3{R_1}\\{R_1} = {R_1} - 7{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&2&0\\0&0&0&1\end{array}} \right]::\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]
Step-2: The pivot elements
Comparing the echelon matrix with original matrix A, we can find the positions of pivot elements:\[M = \left[ {\begin{array}{*{20}{c}}{\left( 1 \right)}&0&{ - 1}&0\\0&{\left( 1 \right)}&2&0\\0&0&0&{\left( 1 \right)}\end{array}} \right]\]So, the pivot columns are column 1, 2 and 4. The
The pivot columns are column 1, 2 and 4
Pivot elements are 1, 5 and 1
We are given with a matrix A\[A = \left[ {\begin{array}{*{20}{l}}1&3&5&7\\3&5&7&9\\5&7&9&1\end{array}} \right]\]We have to Row reduce the matrix to reduced echelon form and find the pivot element.
Step-1: Row-Reduced Echelon form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}1&3&5&7\\3&5&7&9\\5&7&9&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&{ - 8}&{ - 16}&{ - 34}\end{array}} \right]::\left\{ \begin{array}{l}{R_2} = {R_2} - 3{R_1}\\{R_3} = {R_3} - 5{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&{ - 4}&{ - 8}&{ - 17}\end{array}} \right]::\left\{ {{R_3} = \dfrac{{{R_3}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&{ - 4}&{ - 8}&{ - 12}\\0&0&0&{ - 5}\end{array}} \right]::\left\{ {{R_3} = {R_3} - {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&7\\0&1&2&3\\0&0&0&1\end{array}} \right]::\left\{ \begin{array}{l}{R_3} = - \dfrac{{{R_3}}}{5}\\{R_2} = - \dfrac{{{R_2}}}{4}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&5&0\\0&1&2&0\\0&0&0&1\end{array}} \right]::\left\{ \begin{array}{l}{R_2} = {R_2} - 3{R_1}\\{R_1} = {R_1} - 7{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&2&0\\0&0&0&1\end{array}} \right]::\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]
Step-2: The pivot elements
Comparing the echelon matrix with original matrix A, we can find the positions of pivot elements:\[M = \left[ {\begin{array}{*{20}{c}}{\left( 1 \right)}&0&{ - 1}&0\\0&{\left( 1 \right)}&2&0\\0&0&0&{\left( 1 \right)}\end{array}} \right]\]So, the pivot columns are column 1, 2 and 4. The
The pivot columns are column 1, 2 and 4
Pivot elements are 1, 5 and 1