## Elementary Linear Algebra 7th Edition

Published by Cengage Learning

# Chapter 4 - Vector Spaces - 4.7 Coordinates and Change of Basis - 4.7 Exercises - Page 210: 31

#### Answer

$$P^{-1}=\left[ \begin {array}{cccccccc} -24&-10&-29&12 \\ 7&3&7&-3\\ 1&0& 3&-1\\ -2&-1&-2&1\end {array} \right].$$

#### Work Step by Step

Given $$B=\{(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)\}, \\ B^{\prime}= \{(1,3,2,-1),(-2,-5,-5,4),(-1,-2,-2,4),(-2,-3,-5,11) \} .$$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccc} 1&3&2&-1&1&0&0&0\\ -2&-5&-5&4&0&1&0&0\\ -1&-2&-2&4&0&0&1&0 \\ -2&-3&-5&11&0&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination to obtain the transition matrix $$\left[\begin{array}{ll}{I_{4}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccc} 1&0&0&0&-24&-10&-29&12 \\ 0&1&0&0&7&3&7&-3\\ 0&0&1&0&1&0& 3&-1\\ 0&0&0&1&-2&-1&-2&1\end {array} \right] .$$ So, we have $$P^{-1}=\left[ \begin {array}{cccccccc} -24&-10&-29&12 \\ 7&3&7&-3\\ 1&0& 3&-1\\ -2&-1&-2&1\end {array} \right].$$

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