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: 19

Answer

$$ P^{-1}=\left[ \begin {array}{cccc} 2&4\\ -1&3 \end {array} \right] .$$

Work Step by Step

Given $$ B=\{(2,4),(-1,3)\}, B^{\prime}=\{(1,0),(0,1)\} .$$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$ \left[B^{\prime} B\right]=\left[ \begin {array}{cccc} 1&0&2&4\\ 0&1&-1&3 \end {array} \right] .$$ Using Gauss-Jordan elimination, the reduced row echelon form is $$ \left[\begin{array}{ll}{I_{2}} & {P^{-1}}\end{array}\right]=\left[ \begin {array}{cccc} 1&0&2&4\\ 0&1&-1&3 \end {array} \right] .$$ So, we have $$ P^{-1}=\left[ \begin {array}{cccc} 2&4\\ -1&3 \end {array} \right] .$$
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