## 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: 33

#### Answer

$$P^{-1}= \left[ \begin {array}{cccccccccc} 1&0&-\frac{5}{4}&-\frac{3}{4}&0 \\ -\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11} \\ {\frac {5}{11}}&{\frac {3}{22}}&-{ \frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&-\frac{1}{4}&\frac{1}{4}&0 \\ -{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}} &\frac{1}{2}&{\frac {5}{11}}\end {array} \right].$$

#### Work Step by Step

Given $$B=\{(1,0,0,0,0),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1) \} , \\ B^{\prime}= \{(1,2,4,-1,2),(-2,-3,4,2,1),(0,1,2,-2,1),(0,1,2,2,1),(1,-1,0,1,2) \} .$$ To find the transition matrix from $B$ to $B^{\prime}$, we form the matrix $$\left[B^{\prime} B\right]= \left[ \begin {array}{cccccccccc} 1&2&4&-1&2&1&0&0&0&0 \\ -2&-3&4&2&1&0&1&0&0&0\\ 0&1&2&- 2&1&0&0&1&0&0\\ 0&1&2&2&1&0&0&0&1&0 \\ 1&-1&0&1&2&0&0&0&0&1\end {array} \right] .$$ Using Gauss-Jordan elimination to obtain the transition matrix $$\left[\begin{array}{ll}{I_{5}} & {P^{-1}}\end{array}\right]= \left[ \begin {array}{cccccccccc} 1&0&0&0&0&1&0&-\frac{5}{4}&-\frac{3}{4}&0 \\ 0&1&0&0&0&-\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11} \\ 0&0&1&0&0&{\frac {5}{11}}&{\frac {3}{22}}&-{ \frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&0&1&0&0&0&-\frac{1}{4}&\frac{1}{4}&0 \\ 0&0&0&0&1&-{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}} &\frac{1}{2}&{\frac {5}{11}}\end {array} \right] .$$ So, we have $$P^{-1}= \left[ \begin {array}{cccccccccc} 1&0&-\frac{5}{4}&-\frac{3}{4}&0 \\ -\frac{3}{11}&-\frac{2}{11}&{\frac {9}{22}}&\frac{1}{2}&-\frac{1}{11} \\ {\frac {5}{11}}&{\frac {3}{22}}&-{ \frac {19}{44}}&-\frac{1}{4}&-\frac{2}{11}\\ 0&0&-\frac{1}{4}&\frac{1}{4}&0 \\ -{\frac {7}{11}}&-\frac{1}{11}&{\frac {21}{22}} &\frac{1}{2}&{\frac {5}{11}}\end {array} \right].$$

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