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Problems #5,13 pg.222 are graded.
#5, pg.222: Find the coordinate vector [→x]B of →x=[−11] relative to the basis B={[1−2],[3−5]}.
Solution: The coordinates of →x are the weights c1,c2 which are solutions to the vector equation
c1[1−2]+c2[3−5]=[−11].
We can solve this vector equation by row reducing the appropriate augmented matrix, so compute
[13−1−2−51]∼[10201−1].
Hence we see that c1=2 and c2=−1. This implies that the coordinate vector in question is given by
[→x]B=[2−1].
#13, pg.222: The set B={1+t2,t+t2,1+2t+t2} is a basis for P2 (polynomials of degree ≤ 2). Find the coordinate vector of →p(t)=1+4t+7t2 relative to B.
Solution: Our goal is to compute [→p]B=[1+4t+7t2]B (both ways of notating this is are ok). This means we have to solve the following vector equation for the weights c1,c2,c3:
c1(1+t2)+c2(t+t2)+c3(1+2t+t2)=1+4t+7t2.
Some algebraic rearrangement shows that this equation is equivalent to
(c1+c2+c3)t2+(c2+2c3)t+(c1+c3)=1+4t+7t2.
Equating coefficients here leads us to the following system of equations:
{c1+c2+c3=7c2+2c3=4c1+c3=1.
This system can be solved by row reducing the associated augmented matrix, so compute
[111701241011]∼[10020106001−1].
This shows that c1=2,c2=6,c3=−1 and hence
[1+4t+7t2]B=[26−1].
