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 1E from Chapter 2.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
In Exercises 1–6, solve the equation Ax = b by using the LU factorization given for A. In Exercises 1 and 2, also solve Ax = b by ordinary row reduction.
Step-by-Step Solution
Given Information
We are given with following matrices: \[ \begin{array} { l } { \mathbf { A } = \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { - 3 } & { 5 } & { 1 } \\ { 6 } & { - 4 } & { 0 } \end{array} \right] , b = \left[ \begin{array} { c } { - 7 } \\ { 5 } \\ { 2 } \end{array} \right] } \\ { \mathbf { A } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c c c } { 3 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] } \end{array} \] We have to solve the equation $Ax=b$ using the LU factorization. Also solve it using the LU factorization.
Step-1:
LU decomposition of A is given \[ \mathbf { A } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \] With, \[ \mathbf { L } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \text { and } \mathbf { U } = \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \]
Step-2: Solve $Ax=b$
Let, \[ \mathbf { z } = \left[ \begin{array} { l } { z _ { 1 } } \\ { z _ { 2 } } \\ { z _ { 3 } } \end{array} \right] \] Write $\mathbf { U } \mathbf { x } = \mathbf { z }$ \[ \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c } { z _ { 1 } } \\ { z _ { 2 } } \\ { z _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 7 } \\ { 5 } \\ { 2 } \end{array} \right] \] From first Row: $z _ { 1 } = - 7$
From second row: \[ \begin{aligned} - ( - 7 ) + z _ { 2 } & = 5 \\ z _ { 2 } & = - 2 \end{aligned} \] From Third row: \[ \begin{aligned} 2 ( - 7 ) - 5 ( - 2 ) + z _ { 3 } & = 2 \\ - 4 + z _ { 3 } & = 2 \\ z _ { 3 } & = 6 \end{aligned} \] So,
\[ \mathbf { z } = \left[ \begin{array} { r } { - 7 } \\ { - 2 } \\ { 6 } \end{array} \right] \]
Step-3: Solve $Ux=z$
Let, \[ \mathbf { x } = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] \] Then \[ \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 7 } \\ { - 2 } \\ { 6 } \end{array} \right] \] From Third Row: $x _ { 3 } = 6$
From second row: \[ \begin{aligned} - 2 x _ { 2 } - ( - 6 ) & = - 2 \\ - 2 x _ { 2 } & = - 8 \\ x _ { 2 } & = 4 \end{aligned} \] From Third row: \[ \begin{aligned} 3 x _ { 1 } - 7 ( 4 ) - 2 ( - 6 ) & = - 7 \\ 3 x _ { 1 } - 16 & = - 7 \\ 3 x _ { 1 } & = 9 \\ x _ { 1 } & = 3 \end{aligned} \] Therefore,
\[ \mathbf { x } = \left[ \begin{array} { c } { 3 } \\ { 4 } \\ { - 6 } \end{array} \right] \]
Step-4: Row_Reduction Method
The augmented matrix: \[ [ \mathbf { A } | \mathbf { b } ] = \left[ \begin{array} { c c c | c } { 3 } & { - 7 } & { - 2 } & { - 7 } \\ { - 3 } & { 5 } & { 1 } & { 5 } \\ { 6 } & { - 4 } & { 0 } & { 2 } \end{array} \right] \] Row-Reduce the matrix: \[\begin{array}{l} [{\bf{A}}|{\bf{b}}] = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ { - 3}&5&1&5\\ 6&{ - 4}&0&2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ 0&{ - 2}&{ - 1}&{ - 2}\\ 0&{10}&4&{16} \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} = {R_2} + {R_1}}\\ {{R_3} = {R_3} - 2{R_1}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ 0&{ - 2}&{ - 1}&{ - 2}\\ 0&0&{ - 1}&6 \end{array}} \right]::\,\left\{ {{R_3} = {R_3} + 5{R_2}} \right\} \end{array}\]
Step-5: Solve the Equations
From Third Row: $x _ { 3 } = 6$
From second row: \[ \begin{aligned} - 2 x _ { 2 } - ( - 6 ) & = - 2 \\ - 2 x _ { 2 } & = - 8 \\ x _ { 2 } & = 4 \end{aligned} \] From First row: \[ \begin{aligned} 3 x _ { 1 } - 7 ( 4 ) - 2 ( - 6 ) & = - 7 \\ 3 x _ { 1 } - 16 & = - 7 \\ 3 x _ { 1 } & = 9 \\ x _ { 1 } & = 3 \end{aligned} \] Therefore,
\[ \mathbf { x } = \left[ \begin{array} { c } { 3 } \\ { 4 } \\ { - 6 } \end{array} \right] \]
We are given with following matrices: \[ \begin{array} { l } { \mathbf { A } = \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { - 3 } & { 5 } & { 1 } \\ { 6 } & { - 4 } & { 0 } \end{array} \right] , b = \left[ \begin{array} { c } { - 7 } \\ { 5 } \\ { 2 } \end{array} \right] } \\ { \mathbf { A } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c c c } { 3 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] } \end{array} \] We have to solve the equation $Ax=b$ using the LU factorization. Also solve it using the LU factorization.
Step-1:
LU decomposition of A is given \[ \mathbf { A } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \] With, \[ \mathbf { L } = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \text { and } \mathbf { U } = \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \]
Step-2: Solve $Ax=b$
Let, \[ \mathbf { z } = \left[ \begin{array} { l } { z _ { 1 } } \\ { z _ { 2 } } \\ { z _ { 3 } } \end{array} \right] \] Write $\mathbf { U } \mathbf { x } = \mathbf { z }$ \[ \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 1 } & { 1 } & { 0 } \\ { 2 } & { - 5 } & { 1 } \end{array} \right] \left[ \begin{array} { c } { z _ { 1 } } \\ { z _ { 2 } } \\ { z _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 7 } \\ { 5 } \\ { 2 } \end{array} \right] \] From first Row: $z _ { 1 } = - 7$
From second row: \[ \begin{aligned} - ( - 7 ) + z _ { 2 } & = 5 \\ z _ { 2 } & = - 2 \end{aligned} \] From Third row: \[ \begin{aligned} 2 ( - 7 ) - 5 ( - 2 ) + z _ { 3 } & = 2 \\ - 4 + z _ { 3 } & = 2 \\ z _ { 3 } & = 6 \end{aligned} \] So,
\[ \mathbf { z } = \left[ \begin{array} { r } { - 7 } \\ { - 2 } \\ { 6 } \end{array} \right] \]
Step-3: Solve $Ux=z$
Let, \[ \mathbf { x } = \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] \] Then \[ \left[ \begin{array} { c c c } { 3 } & { - 7 } & { - 2 } \\ { 0 } & { - 2 } & { - 1 } \\ { 0 } & { 0 } & { - 1 } \end{array} \right] \left[ \begin{array} { l } { x _ { 1 } } \\ { x _ { 2 } } \\ { x _ { 3 } } \end{array} \right] = \left[ \begin{array} { c } { - 7 } \\ { - 2 } \\ { 6 } \end{array} \right] \] From Third Row: $x _ { 3 } = 6$
From second row: \[ \begin{aligned} - 2 x _ { 2 } - ( - 6 ) & = - 2 \\ - 2 x _ { 2 } & = - 8 \\ x _ { 2 } & = 4 \end{aligned} \] From Third row: \[ \begin{aligned} 3 x _ { 1 } - 7 ( 4 ) - 2 ( - 6 ) & = - 7 \\ 3 x _ { 1 } - 16 & = - 7 \\ 3 x _ { 1 } & = 9 \\ x _ { 1 } & = 3 \end{aligned} \] Therefore,
\[ \mathbf { x } = \left[ \begin{array} { c } { 3 } \\ { 4 } \\ { - 6 } \end{array} \right] \]
Step-4: Row_Reduction Method
The augmented matrix: \[ [ \mathbf { A } | \mathbf { b } ] = \left[ \begin{array} { c c c | c } { 3 } & { - 7 } & { - 2 } & { - 7 } \\ { - 3 } & { 5 } & { 1 } & { 5 } \\ { 6 } & { - 4 } & { 0 } & { 2 } \end{array} \right] \] Row-Reduce the matrix: \[\begin{array}{l} [{\bf{A}}|{\bf{b}}] = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ { - 3}&5&1&5\\ 6&{ - 4}&0&2 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ 0&{ - 2}&{ - 1}&{ - 2}\\ 0&{10}&4&{16} \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{R_2} = {R_2} + {R_1}}\\ {{R_3} = {R_3} - 2{R_1}} \end{array}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 3&{ - 7}&{ - 2}&{ - 7}\\ 0&{ - 2}&{ - 1}&{ - 2}\\ 0&0&{ - 1}&6 \end{array}} \right]::\,\left\{ {{R_3} = {R_3} + 5{R_2}} \right\} \end{array}\]
Step-5: Solve the Equations
From Third Row: $x _ { 3 } = 6$
From second row: \[ \begin{aligned} - 2 x _ { 2 } - ( - 6 ) & = - 2 \\ - 2 x _ { 2 } & = - 8 \\ x _ { 2 } & = 4 \end{aligned} \] From First row: \[ \begin{aligned} 3 x _ { 1 } - 7 ( 4 ) - 2 ( - 6 ) & = - 7 \\ 3 x _ { 1 } - 16 & = - 7 \\ 3 x _ { 1 } & = 9 \\ x _ { 1 } & = 3 \end{aligned} \] Therefore,
\[ \mathbf { x } = \left[ \begin{array} { c } { 3 } \\ { 4 } \\ { - 6 } \end{array} \right] \]