Linear Algebra Lagrange polynomials & Basis

1. Jan 21, 2008

http://img100.imageshack.us/img100/9016/linalggp1.jpg [Broken]
for (a): does that mean i must compute l0(t), l1(t) and l2(t), and i wasn't sure how to do this with the lagrange polynomial formula given, so i found one online and did it, i'm not sure if this is correct, but my l0(t) looks like this:
= (t-a1)(t-a2)/(ao-a1)(ao-a2)
= (t-2)(t-3)/(1-2)(1-3)
=1/2(t-2)(t-3)

for (b) how do i go about proving this?

and also (c) what does it mean to deduce, i have looked up the definition but i still don't understand.

Last edited by a moderator: May 3, 2017
2. Jan 24, 2008

HallsofIvy

Staff Emeritus
Yes, your polynomial for (a) is correct. I'm not sure why you say you "weren't sure how to do it with the given polynomial", it looks like you have done exactly that.
With i= 0, the fraction is $(t- a_1)(t-a_2)/(a_0-a1)(a_0-a2)= (t- 2)(t-3)/(1-2)(1-3)= (t-2)(t-3)/(-1)(-2)= -1/2 (t- 2)(t- 3)$.
With i= 1, the fraction is $(t- a_0)(t-a_2)/(a_1-a_0)(a_1-a_2)= (t-1)(t-3)/(2-1)(2-3)= (t-1)(t-3)/(1)(-1)= -1(t-1)(t-3)$.
With i= 2, the fraction is $(t- a_0)(t-a_1)/(a_2-a_0)(a_2-a_1)= (t-1)(t-2)/(3-1)(3-2)= (t-1)(t-2)/(2)(1)= 1/2(t-1)(t-2)$