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 12E from Chapter 5.4 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
in Exercises, find the B-matrix for the transformation x↦Ax, when = {b1, b2}.
Step-by-Step Solution
Given Information
We are given with the matrix A and vectors in basis B as: \[ \begin{array} { l } { A = \left[ \begin{array} { r r } { - 1 } & { 4 } \\ { - 2 } & { 3 } \end{array} \right] } \\ { \mathbf { b } _ { 1 } = \left[ \begin{array} { c } { 3 } \\ { 2 } \end{array} \right] } \\ { \mathbf { b } _ { 2 } = \left[ \begin{array} { c } { - 1 } \\ { 1 } \end{array} \right] } \end{array} \] We have to find the B-matrix for the transformation \[ \mathbf { x } \mapsto A \mathbf { x } \]
Step-1:
In $A = P D P ^ { - 1 }$, the matrix D represents the B-matrix for the transformation defined by $\mathbf { x } \mapsto A \mathbf { x }$
Step-2:
The matrix P formed by vectors $ b_1$ and $b_2$ \[ P = \left[ \begin{array} { c c } { 3 } & { - 1 } \\ { 2 } & { 1 } \end{array} \right] \]
Step-3:
Inverse of the matrix P is given by: \[\begin{array}{l} {P^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}} d&{ - b}\\ { - c}&a \end{array}} \right]\\ = \dfrac{1}{{(3)(1) - ( - 1)(2)}}{\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]^T}\\ = \dfrac{1}{5}{\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]^T}\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right] \end{array}\]
Step-4: The B-matrix
\[\begin{array}{l} D = {P^{ - 1}}AP\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&4\\ { - 2}&3 \end{array}} \right]\cdot\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 3 + 8}&{1 + 4}\\ { - 6 + 6}&{2 + 3} \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 5&5\\ 0&5 \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} {5 + 0}&{5 + 5}\\ { - 10 + 0}&{ - 10 + 15} \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 5&{10}\\ { - 10}&5 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 2}&1 \end{array}} \right] \end{array}\] Therefore the B-matrix of transformation is given by:
\[B = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 2}&1 \end{array}} \right]\]
We are given with the matrix A and vectors in basis B as: \[ \begin{array} { l } { A = \left[ \begin{array} { r r } { - 1 } & { 4 } \\ { - 2 } & { 3 } \end{array} \right] } \\ { \mathbf { b } _ { 1 } = \left[ \begin{array} { c } { 3 } \\ { 2 } \end{array} \right] } \\ { \mathbf { b } _ { 2 } = \left[ \begin{array} { c } { - 1 } \\ { 1 } \end{array} \right] } \end{array} \] We have to find the B-matrix for the transformation \[ \mathbf { x } \mapsto A \mathbf { x } \]
Step-1:
In $A = P D P ^ { - 1 }$, the matrix D represents the B-matrix for the transformation defined by $\mathbf { x } \mapsto A \mathbf { x }$
Step-2:
The matrix P formed by vectors $ b_1$ and $b_2$ \[ P = \left[ \begin{array} { c c } { 3 } & { - 1 } \\ { 2 } & { 1 } \end{array} \right] \]
Step-3:
Inverse of the matrix P is given by: \[\begin{array}{l} {P^{ - 1}} = \dfrac{1}{{ad - bc}}\left[ {\begin{array}{*{20}{c}} d&{ - b}\\ { - c}&a \end{array}} \right]\\ = \dfrac{1}{{(3)(1) - ( - 1)(2)}}{\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]^T}\\ = \dfrac{1}{5}{\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]^T}\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right] \end{array}\]
Step-4: The B-matrix
\[\begin{array}{l} D = {P^{ - 1}}AP\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 1}&4\\ { - 2}&3 \end{array}} \right]\cdot\left[ {\begin{array}{*{20}{c}} 3&{ - 1}\\ 2&1 \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 3 + 8}&{1 + 4}\\ { - 6 + 6}&{2 + 3} \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 1&1\\ { - 2}&3 \end{array}} \right]\left[ {\begin{array}{*{20}{l}} 5&5\\ 0&5 \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} {5 + 0}&{5 + 5}\\ { - 10 + 0}&{ - 10 + 15} \end{array}} \right]\\ = \dfrac{1}{5}\left[ {\begin{array}{*{20}{c}} 5&{10}\\ { - 10}&5 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 2}&1 \end{array}} \right] \end{array}\] Therefore the B-matrix of transformation is given by:
\[B = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 2}&1 \end{array}} \right]\]