Elementary Linear Algebra 7th Edition

Published by Cengage Learning

Chapter 4 - Vector Spaces - Review Exercises - Page 222: 65

Answer

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

Work Step by Step

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

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