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 2.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
In Exercises, solve the equation Ax = b by using the LU factorization given for A.
Step-by-Step Solution
Given Information
We are given that a matrix and vector \[ A = \left[ \begin{array} { c c c c } { 1 } & { - 2 } & { - 4 } & { - 3 } \\ { 2 } & { - 7 } & { - 7 } & { - 6 } \\ { - 1 } & { 2 } & { 6 } & { 4 } \\ { - 4 } & { - 1 } & { 9 } & { 8 } \end{array} \right] , b = \left[ \begin{array} { l } { 1 } \\ { 7 } \\ { 0 } \\ { 3 } \end{array} \right] \] We have to solve the system M by LU factorization.
Step-1:
The LU factorization of matrix A is given to us: \[\begin{array}{l} L = \left[ {\begin{array}{*{20}{c}} 1&0&0&0\\ 2&1&0&0\\ { - 1}&0&1&0\\ { - 4}&3&{ - 5}&1 \end{array}} \right]\\ \,\,U = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&{ - 3}\\ 0&{ - 3}&1&0\\ 0&0&2&1\\ 0&0&0&1 \end{array}} \right] \end{array}\]
Step-2: Solve Ly=b
\[\begin{array}{l} [L\,\,b] = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 2&1&0&0&7\\ { - 1}&0&1&0&0\\ { - 4}&3&{ - 5}&1&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&3&{ - 5}&1&7 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} - 2{R_1}\\ {R_3} = {R_3} + {R_1}\\ {R_4} = {R_4} + 4{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&0&{ - 5}&1&{ - 8} \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 3{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} + 5{R_3}} \right\} \end{array}\] So, \[ y = \left[ \begin{array} { c } { 1 } \\ { 5 } \\ { 1 } \\ { - 3 } \end{array} \right] \]
Step-3: Solve Ux=y
\[\begin{array}{l} [U\,\,y] = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&{ - 3}&1\\ 0&{ - 3}&1&0&5\\ 0&0&2&1&1\\ 0&0&0&1&{ - 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&0&{ - 8}\\ 0&{ - 3}&1&0&5\\ 0&0&2&0&4\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_1} = {R_1} + 3{R_4}\\ {R_3} = {R_3} - {R_4} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&0&{ - 8}\\ 0&{ - 3}&1&0&5\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_3} = \dfrac{{{R_3}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&0&0&0\\ 0&{ - 3}&0&0&3\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_1} = {R_1} + 4{R_3}\\ {R_2} = {R_2} - {R_3} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&0&0&0\\ 0&1&0&0&{ - 1}\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_2} = - \dfrac{1}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{ - 2}\\ 0&1&0&0&{ - 1}\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 2{R_2}} \right\}\\ = \left[ {I\,\,\,\,\,x} \right] \end{array}\] Therefore, the solution of $Ax=b$ is:
\[ \mathrm { x } = \left[ \begin{array} { c } { - 2 } \\ { - 1 } \\ { 2 } \\ { - 3 } \end{array} \right] \]
We are given that a matrix and vector \[ A = \left[ \begin{array} { c c c c } { 1 } & { - 2 } & { - 4 } & { - 3 } \\ { 2 } & { - 7 } & { - 7 } & { - 6 } \\ { - 1 } & { 2 } & { 6 } & { 4 } \\ { - 4 } & { - 1 } & { 9 } & { 8 } \end{array} \right] , b = \left[ \begin{array} { l } { 1 } \\ { 7 } \\ { 0 } \\ { 3 } \end{array} \right] \] We have to solve the system M by LU factorization.
Step-1:
The LU factorization of matrix A is given to us: \[\begin{array}{l} L = \left[ {\begin{array}{*{20}{c}} 1&0&0&0\\ 2&1&0&0\\ { - 1}&0&1&0\\ { - 4}&3&{ - 5}&1 \end{array}} \right]\\ \,\,U = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&{ - 3}\\ 0&{ - 3}&1&0\\ 0&0&2&1\\ 0&0&0&1 \end{array}} \right] \end{array}\]
Step-2: Solve Ly=b
\[\begin{array}{l} [L\,\,b] = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 2&1&0&0&7\\ { - 1}&0&1&0&0\\ { - 4}&3&{ - 5}&1&3 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&3&{ - 5}&1&7 \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_2} = {R_2} - 2{R_1}\\ {R_3} = {R_3} + {R_1}\\ {R_4} = {R_4} + 4{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&0&{ - 5}&1&{ - 8} \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - 3{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&1\\ 0&1&0&0&5\\ 0&0&1&0&1\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} + 5{R_3}} \right\} \end{array}\] So, \[ y = \left[ \begin{array} { c } { 1 } \\ { 5 } \\ { 1 } \\ { - 3 } \end{array} \right] \]
Step-3: Solve Ux=y
\[\begin{array}{l} [U\,\,y] = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&{ - 3}&1\\ 0&{ - 3}&1&0&5\\ 0&0&2&1&1\\ 0&0&0&1&{ - 3} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&0&{ - 8}\\ 0&{ - 3}&1&0&5\\ 0&0&2&0&4\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_1} = {R_1} + 3{R_4}\\ {R_3} = {R_3} - {R_4} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{ - 4}&0&{ - 8}\\ 0&{ - 3}&1&0&5\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_3} = \dfrac{{{R_3}}}{2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&0&0&0\\ 0&{ - 3}&0&0&3\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ \begin{array}{l} {R_1} = {R_1} + 4{R_3}\\ {R_2} = {R_2} - {R_3} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&0&0&0\\ 0&1&0&0&{ - 1}\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_2} = - \dfrac{1}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&0&0&{ - 2}\\ 0&1&0&0&{ - 1}\\ 0&0&1&0&2\\ 0&0&0&1&{ - 3} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 2{R_2}} \right\}\\ = \left[ {I\,\,\,\,\,x} \right] \end{array}\] Therefore, the solution of $Ax=b$ is:
\[ \mathrm { x } = \left[ \begin{array} { c } { - 2 } \\ { - 1 } \\ { 2 } \\ { - 3 } \end{array} \right] \]