Polynomial coordinates
Answer:
\[p\]B = [16, -25, 6]
Explanation:
Okay, let's go through this step-by-step:
The three basis vectors for the basis B are:
b1 = 1
b2 = 1 - t
b3 = 2 - 4t + t^2
To find the coordinate vector of p(t) = 16 - 25t + 6t^2 relative to this basis B, we need to solve the equation:
p(t) = c1b1 + c2b2 + c3b3
Plugging in the expressions:
16 - 25t + 6t^2 = c1(1) + c2(1 - t) + c3(2 - 4t + t^2)
Equating coefficients gives:
c1 = 16
c2 = -25
c3 = 6
Therefore, the coordinate vector of p(t) relative to the basis B is:
\[p\]B = [16, -25, 6]