(i):
The coordinates of p(x)∈ P2 with respect to the basis B is [p(x)]B=⎣⎡1−13⎦⎤
∴p(x)=1(1−x)−1(2+x)+3(3−x+x2)=(1−2+9).1+(−1−1−3)x+3.x2=8−5x+3x2
(ii):
Let 1−x=a.1+b(2+x)+c(1+x−x2)
⇒1−x=(a+2b+c)+(b+c)x−cx2
On comparing, we get,
a+2b+c=1b+c=−1−c=0⇒c=0,b=−1,a=3
So, 1−x=3(1)−1(2+x)+0.(1+x−x2) ...(i)
Also, 2+x=0(1)+1(2+x)+0.(1+x−x2) ...(ii)
Also, 3−x+x2=4(1)+0(2+x)+(−1).(1+x−x2) ...(iii)
From (i), (ii), (iii), we get the transition matrix from B to C as:
CMB=⎣⎡3−1001040−1⎦⎤
(iii):
From part (i), we have p(x)=8−5x+3x2
Let 8−5x+3x2=p(1)+q(2+x)+r(1+x−x2)
⇒8−5x+3x2=(p+2q+r)+(q+r)x−rx2
On comparing, we get,
p+2q+r=8q+r=−5−r=3⇒r=−3,q=−2,p=15
So, we get, p(x)=15(1)−2(2+x)−3(1+x−x2)
So, the coordinates of p(x) w.r.t basis C is:
[p(x)]C=⎣⎡15−2−3⎦⎤
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