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: 21

Answer

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

Work Step by Step

Given $$ B=\{(1,0,0),(0,1,0),(0,0,1)\} , \quad B^{\prime}=\{(1,0,0),(0,2,8),(6,0,12)\} $$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$ \left[B^{\prime} B\right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&2&8&0 &1&0\\ 6&0&12&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination to obtain the transition matrix $$ \left[\begin{array}{ll}{I_{3}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccc} 1&0&0&1&0&0\\ 0&1&0&2 &\frac{1}{2}&-\frac{1}{3}\\ 0&0&1&-\frac{1}{2}&0&\frac{1}{2}\end {array} \right] .$$ So, we have $$ P^{-1}=\left[ \begin {array}{cccccc} 1&0&0\\ 2 &\frac{1}{2}&-\frac{1}{3}\\ -\frac{1}{2}&0&\frac{1}{2}\end {array} \right] $$
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