Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 4 - Vector Spaces - 4.7 Coordinates and Change of Basis - 4.7 Exercises - Page 210: 28


$$ P^{-1}=\left[ \begin {array}{cccccc} 0&\frac{1}{3}&-\frac{1}{3}\\ -\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ \frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right]. $$

Work Step by Step

Given $$ B=\{(1,0,0),(0,1,0),(0,0,1)\}, \quad B^{\prime}=\{(2,-1,4),(0,2,1),(-3,2,1)\}. $$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$ \left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 2&-1&4&1&0&0\\ 0&2&1&0 &1&0\\ -3&2&1&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination to obtain the transition matrix $$ \left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&0&\frac{1}{3}&-\frac{1}{3}\\ 0& 1&0&-\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ 0&0&1&\frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right] .$$ So, we have $$ P^{-1}=\left[ \begin {array}{cccccc} 0&\frac{1}{3}&-\frac{1}{3}\\ -\frac{1}{9}&\frac {14}{27}&-\frac {2}{27}\\ \frac{2}{9}&-\frac{1}{27}&{\frac {4}{27}}\end {array} \right]. $$
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