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 10E from Chapter 4.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 10E
Chapter:
Problem:
Let = {b1, b2} and = {c1, c2} be bases for ℝ2. In each exercise, find the change-of-coordinates matrix from to and the change-of-coordinates matrix from to .
Step-by-Step Solution
Given Information
We are given that ${\rm{B}} = \left\{ {{{\rm{b}}_1},{{\rm{b}}_2}} \right\}$ and ${\rm{C}} = \left\{ {{{\rm{c}}_1},{{\rm{c}}_2}} \right\}$ be bases for a vector space V. We are also given that \[ \mathbf { b } _ { 1 } = \left[ \begin{array} { r } { 7 } \\ { - 2 } \end{array} \right] , \mathbf { b } _ { 2 } = \left[ \begin{array} { r } { 2 } \\ { - 1 } \end{array} \right] , \mathbf { c } _ { 1 } = \left[ \begin{array} { l } { 4 } \\ { 1 } \end{array} \right] , \mathbf { c } _ { 2 } = \left[ \begin{array} { l } { 5 } \\ { 2 } \end{array} \right] \] We have to find the change-of-coordinate matrix from B to C and also from C to B.
Step-1: Row Reduced augmented matrix
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{c_1}}&{{c_2}}&{{b_1}}&{{b_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&5&7&2\\ 1&2&{ - 2}&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 1&2&{ - 2}&{ - 1} \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{4}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 0&{\dfrac{3}{4}}&{ - \dfrac{{15}}{4}}&{ - \dfrac{3}{2}} \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} - {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 0&1&{ - 5}&{ - 2} \end{array}} \right]::\,\,\left\{ {{R_2} = \dfrac{4}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&8&3\\ 0&1&{ - 5}&{ - 2} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - \dfrac{5}{4}{R_2}} \right\} \end{array}\] Therefore, the change of matrix from B to C is \[{P_{B \to C}} = \left[ {\begin{array}{*{20}{c}} 8&3\\ { - 5}&{ - 2} \end{array}} \right]\]
Step-2: Change of coordinate Matrix from C to B
Take inverse of matrix P \[\begin{array}{l} {P_{C \to B}} = {\left[ {{P_{B \to C}}} \right]^{ - 1}}\\ = {\left[ {\begin{array}{*{20}{c}} 8&3\\ { - 5}&{ - 2} \end{array}} \right]^{ - 1}}\\ = \dfrac{1}{{8( - 2) - ( - 5)3}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \dfrac{1}{{ - 16 + 15}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - 5}&{ - 8} \end{array}} \right] \end{array}\] Therefore change of coordinate matrix from C to B is:
\[{P_{C \to B}} = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - 5}&{ - 8} \end{array}} \right]\]
We are given that ${\rm{B}} = \left\{ {{{\rm{b}}_1},{{\rm{b}}_2}} \right\}$ and ${\rm{C}} = \left\{ {{{\rm{c}}_1},{{\rm{c}}_2}} \right\}$ be bases for a vector space V. We are also given that \[ \mathbf { b } _ { 1 } = \left[ \begin{array} { r } { 7 } \\ { - 2 } \end{array} \right] , \mathbf { b } _ { 2 } = \left[ \begin{array} { r } { 2 } \\ { - 1 } \end{array} \right] , \mathbf { c } _ { 1 } = \left[ \begin{array} { l } { 4 } \\ { 1 } \end{array} \right] , \mathbf { c } _ { 2 } = \left[ \begin{array} { l } { 5 } \\ { 2 } \end{array} \right] \] We have to find the change-of-coordinate matrix from B to C and also from C to B.
Step-1: Row Reduced augmented matrix
\[\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {{c_1}}&{{c_2}}&{{b_1}}&{{b_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4&5&7&2\\ 1&2&{ - 2}&{ - 1} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 1&2&{ - 2}&{ - 1} \end{array}} \right]::\,\,\left\{ {{R_1} = \dfrac{{{R_1}}}{4}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 0&{\dfrac{3}{4}}&{ - \dfrac{{15}}{4}}&{ - \dfrac{3}{2}} \end{array}} \right]::\,\,\left\{ {{R_2} = {R_2} - {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&{\dfrac{5}{4}}&{\dfrac{7}{4}}&{\dfrac{1}{2}}\\ 0&1&{ - 5}&{ - 2} \end{array}} \right]::\,\,\left\{ {{R_2} = \dfrac{4}{3}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} 1&0&8&3\\ 0&1&{ - 5}&{ - 2} \end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} - \dfrac{5}{4}{R_2}} \right\} \end{array}\] Therefore, the change of matrix from B to C is \[{P_{B \to C}} = \left[ {\begin{array}{*{20}{c}} 8&3\\ { - 5}&{ - 2} \end{array}} \right]\]
Step-2: Change of coordinate Matrix from C to B
Take inverse of matrix P \[\begin{array}{l} {P_{C \to B}} = {\left[ {{P_{B \to C}}} \right]^{ - 1}}\\ = {\left[ {\begin{array}{*{20}{c}} 8&3\\ { - 5}&{ - 2} \end{array}} \right]^{ - 1}}\\ = \dfrac{1}{{8( - 2) - ( - 5)3}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \dfrac{1}{{ - 16 + 15}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}\\ 5&8 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - 5}&{ - 8} \end{array}} \right] \end{array}\] Therefore change of coordinate matrix from C to B is:
\[{P_{C \to B}} = \left[ {\begin{array}{*{20}{c}} 2&3\\ { - 5}&{ - 8} \end{array}} \right]\]