the set B= {1+t^2, -t + t^2, 1 + 2t + t^2} is a basis for P2. what is the coordinate vector of p(t) = 5 - t +10t^2 relative to B.

Answer 1

5 - t + 10t² = a(1 + t²) + b(-t + t²) + c(1 + 2t + t²)

= (a + c) + (-b + 2c)t + (a + b + c)t²

So, a + c = 5

-b + 2c = -1

a + b + c = 10

From the first equation, c = 5 - a.

So, -b + 2(5 - a) = -1 and a + b + (5 - a) = 10.

b = 5, a = 3 and c = 2

Coordinate vector is <a, b, c> = <3, 5, 2>

In other words, 5 - t + 10t² = 3(1 + t²) + 5(-t + t²) + 2(1 + 2t + t²).