Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 4 - Vector Spaces - 4.7 Change of Basis - Problems - Page 318: 22

Answer

$[-4+x-6x^2]_C=(-1,3,-3)\\ [6-2x]_C=(2,11,-6)\\ [-6-2x+4x^2]_C=(2,-1,-2) \\ P_{C \leftarrow B}=\begin{bmatrix} -1& 11 & 2\\ 3 & 5 & -1\\ -3 & -6 & -2 \end{bmatrix}$

Work Step by Step

We know that: $a_1(1-x+3x^2)+a_2(2)+a_3(3+x^2)=-4+x-6x^2$ $b_1(1-x+3x^2)+b_2(2)+b_3(3+x^2)=-6-2x$ $c_1(1-x+3x^2)+c_2(2)+c_3(3+x^2)=-6-2x+x^2$ To find $[-4+x-6x^2]_C$: $a_1(1-x+3x^2)+a_2(2)+a_3(3+x^2)\\ =a_1-a_1x+3a_1x^2+2a_2+3a_3+a_3x^2\\ =(a_1+2a_2+3a_3)+x(-a_1)+x^2(3a_1+a_3)\\ =-4+x-6x^2$ Thus $a_1+2a_2=-4\\ -a_1=1 \rightarrow a_1=-1\\ 3a_1+a_3=-6 \rightarrow -3+a_3=-6 \rightarrow a_3=-3\\ -1+2a_2+3(-3)=-4\rightarrow a_2=3$ To find $[-6-2x]_{C}:$ $b_1(1-x+3x^2)+b_2(2)+b_3(3+x^2)\\ =b_1-b_1+3b_1x^2+2b_2+3b_3+b_3x^2\\ =(b_1+2b_2+3b_3)+x(b_1)+x^2(3b_1+b_3)\\ =-6-2x$ Thus $b_1+2b_2+3b_3=-6\\ -b_1=-2 \rightarrow b_1=2\\ 3b_1+b_3=0 \rightarrow b_3=-6 \\ 2+2b_2+3(-6)=0 \rightarrow b_2=11$ To find $[-6-2x+4x^2]_{C}:$ $c_1(1-x+3x^2)+c_2(2)+c_3(3+x^2)\\ =c_1-c_1+3c_1x^2+2c_2+3c_3+c_3x^2\\ =(c_1+2c_2+3c_3)+x(c_1)+x^2(3c_1+c_3)\\ =-6-2x+4x^2$ Thus $c_1+2c_2+3c_3=-6\\ -c_1=-2 \rightarrow c_1=2\\ 3c_1+c_3=4 \rightarrow c_3=-2 \\ 2+2c_2-6=-6 \rightarrow c_2=-1$ Hence we have: $[-4+x-6x^2]_C=(-1,3,-3)\\ [6-2x]_C=(2,11,-6)\\ [-6-2x+4x^2]_C=(2,-1,-2) \\ P_{C \leftarrow B}=\begin{bmatrix} -1& 11 & 2\\ 3 & 5 & -1\\ -3 & -6 & -2 \end{bmatrix}$
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